Cambridge IGCSE · Thinka-original Practice Paper

2023 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Thinka Jun 2023 (V3) Cambridge International A Level-Style Mock — Biology (0610)

160 marks180 mins2023
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V3) Cambridge International A Level Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

Paper 23

Answer all 40 multiple-choice questions.
33 Question · 33 marks
Question 1 · multiple-choice
1 marks
Under which set of environmental conditions will a leafy shoot experience the highest rate of transpiration?
  1. A.High humidity, high temperature, high wind speed
  2. B.High humidity, low temperature, low wind speed
  3. C.Low humidity, high temperature, high wind speed
  4. D.Low humidity, low temperature, low wind speed Gold standard conditions for high transpiration rate
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Worked solution

Low humidity increases the water potential gradient between the inside of the leaf and the surrounding air. High temperature increases the kinetic energy of water molecules, increasing evaporation. High wind speed removes saturated air from the leaf surface, maintaining a steep concentration gradient. Therefore, these combined conditions lead to the highest transpiration rate.

Marking scheme

Award 1 mark for identifying option C as the correct combination.
Question 2 · multiple-choice
1 marks
Which row correctly identifies the state of the heart valves when the ventricles are contracting?
  1. A.Atrioventricular valves closed, semilunar valves closed
  2. B.Atrioventricular valves closed, semilunar valves open
  3. C.Atrioventricular valves open, semilunar valves closed
  4. D.Atrioventricular valves open, semilunar valves open
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Worked solution

During ventricular contraction (systole), the high pressure of blood in the ventricles forces the atrioventricular valves (bicuspid and tricuspid) to close, preventing backflow of blood into the atria. Simultaneously, this pressure forces the semilunar valves open so blood can flow out into the aorta and pulmonary artery.

Marking scheme

Award 1 mark for identifying option B as the correct state of both valves during ventricular contraction.
Question 3 · multiple-choice
1 marks
Which row correctly identifies a substance that is filtered from the blood into the Bowman's capsule and its status of reabsorption in a healthy kidney?
  1. A.Substance filtered: Glucose and urea; Reabsorption: Glucose is completely reabsorbed
  2. B.Substance filtered: Proteins and glucose; Reabsorption: Proteins are completely reabsorbed
  3. C.Substance filtered: Red blood cells and salts; Reabsorption: Salts are completely reabsorbed
  4. D.Substance filtered: Urea and proteins; Reabsorption: Urea is completely reabsorbed
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Worked solution

Both glucose and urea are small molecules that are forced out of the blood plasma through the glomerulus into the Bowman's capsule during ultrafiltration. In a healthy kidney, all of the glucose is selectively reabsorbed back into the blood in the proximal convoluted tubule, whereas urea is not reabsorbed and is excreted in urine.

Marking scheme

Award 1 mark for identifying option A as the correct combination of filtered and reabsorbed substances.
Question 4 · multiple-choice
1 marks
A student performs food tests on an unknown liquid sample. The results are: Biuret test turns purple; Ethanol emulsion test forms a cloudy white emulsion; Iodine test remains yellow-brown. Which biological molecules are present in this sample?
  1. A.Protein and starch
  2. B.Protein and lipid
  3. C.Reducing sugar and lipid
  4. D.Starch and lipid
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Worked solution

A positive Biuret test (purple) indicates the presence of protein. A positive ethanol emulsion test (cloudy white emulsion) indicates the presence of lipids. A negative iodine test (remaining yellow-brown) indicates that starch is absent. Therefore, protein and lipid are present.

Marking scheme

Award 1 mark for selecting option B.
Question 5 · multiple-choice
1 marks
Cylinders of potato tissue of equal initial length were placed in sucrose solutions of different concentrations. After two hours, cylinder P decreased in length, while cylinder Q increased in length. Which statement about water potential is correct?
  1. A.The water potential of solution P was higher than that of solution Q.
  2. B.The water potential of the cells of cylinder P was lower than that of solution P.
  3. C.The water potential of solution Q was higher than that of the cells of cylinder Q.
  4. D.The water potential of the cells of cylinder Q was higher than that of solution Q.
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Worked solution

Because cylinder Q increased in length, water must have entered its cells by osmosis. Water moves down a water potential gradient from a region of higher water potential to a region of lower water potential. Therefore, solution Q must have had a higher water potential than the cells of cylinder Q.

Marking scheme

Award 1 mark for identifying option C as the only correct statement about water potential gradients.
Question 6 · multiple-choice
1 marks
In a species of plant, the allele for red flowers (R) is dominant to the allele for white flowers (r). A heterozygous red-flowered plant is crossed with a white-flowered plant. What is the expected ratio of phenotypes in the offspring?
  1. A.1 red-flowered : 1 white-flowered
  2. B.3 red-flowered : 1 white-flowered
  3. C.All red-flowered
  4. D.All white-flowered
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Worked solution

The heterozygous red-flowered plant has the genotype Rr. The white-flowered plant must be homozygous recessive (rr). The cross Rr x rr produces offspring with genotypes Rr and rr in a 1:1 ratio. Phenotypically, this corresponds to 50% red-flowered plants and 50% white-flowered plants, which is a 1:1 ratio.

Marking scheme

Award 1 mark for correctly identifying the 1:1 phenotypic ratio (option A).
Question 7 · multiple-choice
1 marks
Which sequence of events is correct during the production of human insulin by genetically modified bacteria?
  1. A.extract human gene -> insert gene into bacterial plasmid -> insert plasmid into bacterium -> replicate bacterium in fermenter
  2. B.insert plasmid into bacterium -> extract human gene -> insert gene into bacterial plasmid -> replicate bacterium in fermenter
  3. C.replicate bacterium in fermenter -> extract human gene -> insert gene into bacterial plasmid -> insert plasmid into bacterium
  4. D.extract human gene -> replicate bacterium in fermenter -> insert gene into bacterial plasmid -> insert plasmid into bacterium
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Worked solution

The correct sequence starts with isolating and extracting the human insulin gene. This gene is then inserted into a bacterial plasmid (vector) using enzymes. The recombinant plasmid is introduced into a bacterium, which is then cultured in a fermenter to replicate and express the insulin protein at a large scale.

Marking scheme

Award 1 mark for choosing option A as the correct logical flow of genetic engineering steps.
Question 8 · multiple-choice
1 marks
Which statement best describes the effect of increasing the temperature from 20 to 80 degrees Celsius on an enzyme-catalyzed reaction?
  1. A.The rate of reaction increases continuously because the kinetic energy of the molecules increases.
  2. B.The rate of reaction decreases initially and then increases as the enzyme adapts to the higher temperature.
  3. C.The rate of reaction increases up to the optimum temperature, and then decreases as the shape of the enzyme's active site is altered.
  4. D.The rate of reaction decreases continuously because high temperatures always destroy the substrate molecules.
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Worked solution

As temperature increases up to the optimum, the rate of reaction increases because molecules have more kinetic energy, leading to more frequent successful collisions between enzyme and substrate. Above the optimum temperature, the high temperature disrupts the chemical bonds maintaining the shape of the enzyme's active site, denaturing it so the substrate can no longer fit.

Marking scheme

Award 1 mark for selecting option C.
Question 9 · multiple-choice
1 marks
A student set up a potometer to measure the rate of water uptake of a leafy shoot. She wants to investigate the effect of humidity. How can she increase the humidity around the shoot, and what will be the effect on the rate of movement of the bubble in the capillary tube?
  1. A.Blow a fan across the shoot; the bubble will move faster.
  2. B.Cover the shoot with a clear plastic bag; the bubble will move slower.
  3. C.Cover the shoot with a clear plastic bag; the bubble will move faster.
  4. D.Blow a fan across the shoot; the bubble will move slower.
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Worked solution

Covering the shoot with a plastic bag traps transpired water vapor, which increases the humidity around the leaves. This reduces the water potential gradient between the inside of the leaf and the external air, thereby reducing the rate of transpiration and water uptake. As a result, the bubble in the capillary tube moves slower.

Marking scheme

1 mark: Identify that covering the shoot with a clear plastic bag increases humidity and results in the bubble moving slower.
Question 10 · multiple-choice
1 marks
During ventricular contraction (systole) in the human heart, what is the state of the valves?
  1. A.Valves between atria and ventricles are open; valves in the pulmonary artery and aorta are closed.
  2. B.Valves between atria and ventricles are open; valves in the pulmonary artery and aorta are open.
  3. C.Valves between atria and ventricles are closed; valves in the pulmonary artery and aorta are open.
  4. D.Valves between atria and ventricles are closed; valves in the pulmonary artery and aorta are closed.
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Worked solution

During ventricular contraction, high pressure is generated in the ventricles. To prevent blood from flowing backward into the atria, the valves between the atria and ventricles close. Simultaneously, the high pressure forces open the semilunar valves in the pulmonary artery and aorta, allowing blood to flow into these arteries.

Marking scheme

1 mark: Correct state of both types of valves (closed atrioventricular valves and open arterial valves).
Question 11 · multiple-choice
1 marks
Which comparison of the blood entering a healthy kidney via the renal artery and leaving it via the renal vein is correct?
  1. A.The blood in the renal vein contains more urea, more glucose, and more oxygen.
  2. B.The blood in the renal vein contains less urea, less glucose, and less oxygen.
  3. C.The blood in the renal vein contains less urea, more glucose, and more oxygen.
  4. D.The blood in the renal vein contains more urea, less glucose, and less oxygen.
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Worked solution

The kidney filters out urea from the blood, so the renal vein has a lower concentration of urea. Kidney cells are metabolically active and carry out aerobic respiration, which consumes glucose and oxygen, so the renal vein also has less glucose and less oxygen than the renal artery.

Marking scheme

1 mark: Identify that the renal vein contains less urea, less glucose, and less oxygen compared to the renal artery.
Question 12 · multiple-choice
1 marks
A solution of an unknown food substance was tested with different reagents. The results were as follows: Iodine test was blue-black; Benedict's test (heated) remained blue; Biuret test was purple; Ethanol emulsion test remained clear and colourless. Which biological molecules are present in this food substance?
  1. A.Reducing sugar and protein only
  2. B.Starch and lipid only
  3. C.Reducing sugar, protein, and lipid
  4. D.Starch and protein only
Show answer & marking scheme

Worked solution

A positive iodine test (blue-black) indicates the presence of starch. A negative Benedict's test (remained blue) indicates the absence of reducing sugars. A positive Biuret test (purple) indicates the presence of protein. A negative ethanol emulsion test (clear and colourless) indicates the absence of lipids. Therefore, only starch and protein are present.

Marking scheme

1 mark: Correctly identify that only starch and protein are present based on the food test results.
Question 13 · multiple-choice
1 marks
Four cubes of agar gel containing a pH indicator are placed into a beaker of dilute hydrochloric acid. The acid diffuses into the cubes, turning them from pink to colourless. The side lengths of the cubes are: Cube 1: 0.5 cm; Cube 2: 1.0 cm; Cube 3: 2.0 cm; Cube 4: 3.0 cm. Which cube will take the longest time to become completely colourless?
  1. A.Cube 1 (0.5 cm side length)
  2. B.Cube 4 (3.0 cm side length)
  3. C.Cube 2 (1.0 cm side length)
  4. D.Cube 3 (2.0 cm side length)
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Worked solution

Cube 4 has the largest dimensions (3.0 cm). While the rate of diffusion remains the same, the distance the acid needs to travel to reach the center of Cube 4 is the greatest due to its lower surface-area-to-volume ratio. Therefore, it will take the longest time to become completely colourless.

Marking scheme

1 mark: Correctly identify Cube 4 as taking the longest time to turn completely colourless.
Question 14 · multiple-choice
1 marks
Equal-sized cylinders of potato tissue were placed in sucrose solutions of different concentrations: P, Q, R, and S. After two hours, the change in length of each cylinder was recorded: Cylinder in P increased in length; Cylinder in Q decreased in length; Cylinder in R showed no change in length; Cylinder in S decreased significantly in length. Which sequence shows the sucrose solutions ordered from lowest concentration of sucrose to highest concentration of sucrose?
  1. A.P -> R -> Q -> S
  2. B.S -> Q -> R -> P
  3. C.P -> Q -> R -> S
  4. D.R -> P -> Q -> S
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Worked solution

A low sucrose concentration has a high water potential, so water enters the potato by osmosis, causing it to increase in length (P). A solution with the same water potential as the cell sap results in no net movement of water and no change in length (R). High sucrose concentrations have lower water potential, drawing water out of the cells and causing them to shrink. The higher the concentration, the more the cylinder shrinks (Q decreased, S decreased significantly). Therefore, the correct order from lowest to highest sucrose concentration is P -> R -> Q -> S.

Marking scheme

1 mark: Identify the correct sequence of solutions from lowest to highest concentration (P -> R -> Q -> S).
Question 15 · multiple-choice
1 marks
In a certain plant species, flower colour is determined by two codominant alleles: \(C^R\) (red) and \(C^W\) (white). Heterozygous individuals (\(C^R C^W\)) produce pink flowers. If a pink-flowered plant is crossed with a white-flowered plant, what is the expected ratio of phenotypes in the offspring?
  1. A.100% pink
  2. B.50% red, 50% white
  3. C.50% pink, 50% white
  4. D.25% red, 50% pink, 25% white
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Worked solution

Crossing a pink-flowered plant (\(C^R C^W\)) with a white-flowered plant (\(C^W C^W\)) yields offspring with genotypes \(C^R C^W\) (pink) and \(C^W C^W\) (white) in a 1:1 ratio. Thus, 50% of the offspring are expected to have pink flowers, and 50% are expected to have white flowers.

Marking scheme

1 mark: Correct phenotypic ratio (50% pink, 50% white) determined from the genetic cross.
Question 16 · multiple-choice
1 marks
Which of the following is a key ecological reason for conserving endangered animal species in their natural habitats?
  1. A.To increase the potential for domesticating wild animals.
  2. B.To reduce genetic diversity within the species population.
  3. C.To allow natural ecosystems to be replaced by agricultural land.
  4. D.To maintain biodiversity and preserve food webs.
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Worked solution

Conserving endangered species helps maintain overall biodiversity, which is vital for the stability of ecosystems and the food webs within them. Options A, B, and C do not represent valid ecological reasons or aims of conservation.

Marking scheme

1 mark: Identify the maintenance of biodiversity and preservation of food webs as the correct ecological reason for conservation.
Question 17 · multiple-choice
1 marks
Under which combination of environmental conditions will the rate of transpiration from a leafy shoot be lowest?
  1. A.Dry and breezy
  2. B.Dry and still
  3. C.Humid and breezy
  4. D.Humid and still
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Worked solution

High humidity decreases the water potential gradient between the air spaces inside the leaf and the external atmosphere, which slows the rate of diffusion of water vapor. Still air allows a boundary layer of humid air to build up around the stomata, further reducing this concentration gradient. Therefore, transpiration is slowest in humid and still conditions.

Marking scheme

1 mark for selecting the correct combination of conditions (humid and still).
Question 18 · multiple-choice
1 marks
Which row correctly identifies the chamber of the heart that pumps deoxygenated blood to the lungs and the main blood vessel that carries oxygenated blood to the rest of the body?
  1. A.Left ventricle | Aorta
  2. B.Right ventricle | Aorta
  3. C.Left atrium | Pulmonary artery
  4. D.Right atrium | Pulmonary vein
Show answer & marking scheme

Worked solution

The right ventricle of the heart pumps deoxygenated blood into the pulmonary artery to be oxygenated in the lungs. The left ventricle pumps oxygenated blood into the aorta, which is the main artery carrying oxygenated blood to the rest of the body.

Marking scheme

1 mark for identifying the right ventricle as the chamber and the aorta as the blood vessel.
Question 19 · multiple-choice
1 marks
In the human kidney, which structures are responsible for ultrafiltration and the selective reabsorption of glucose?
  1. A.Glomerulus | Proximal convoluted tubule
  2. B.Bowman's capsule | Loop of Henle
  3. C.Collecting duct | Proximal convoluted tubule
  4. D.Glomerulus | Collecting duct
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Worked solution

Ultrafiltration occurs under high pressure in the glomerulus, where small molecules are forced out of the blood into the Bowman's capsule. Selective reabsorption of useful molecules, such as all glucose, occurs as the filtrate passes through the proximal convoluted tubule.

Marking scheme

1 mark for identifying the glomerulus as the site of ultrafiltration and the proximal convoluted tubule as the site of selective reabsorption of glucose.
Question 20 · multiple-choice
1 marks
A student tests a liquid food sample and records the following results: Iodine test remains orange-brown; Biuret test turns purple; Ethanol emulsion test remains clear; Benedict's test turns red after heating. Which nutrients are present in the food sample?
  1. A.Starch and protein
  2. B.Reducing sugar and protein
  3. C.Protein and lipid
  4. D.Starch and reducing sugar
Show answer & marking scheme

Worked solution

A purple color in the Biuret test indicates the presence of protein. A red color in the Benedict's test after heating indicates a high concentration of reducing sugars. The negative results for the iodine test (remains orange-brown) and the ethanol emulsion test (remains clear) mean starch and lipids are absent.

Marking scheme

1 mark for correctly identifying the presence of reducing sugar and protein based on the food test results.
Question 21 · multiple-choice
1 marks
Which changes would cause the greatest increase in the rate of diffusion of oxygen across a cell membrane?
  1. A.Decreasing the concentration gradient and increasing the surface area
  2. B.Increasing the temperature and decreasing the surface area
  3. C.Increasing the concentration gradient and increasing the temperature
  4. D.Increasing the diffusion distance and increasing the temperature
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Worked solution

The rate of diffusion is increased by a steeper concentration gradient, which provides a stronger driving force for net movement, and a higher temperature, which increases the kinetic energy and speed of the diffusing particles.

Marking scheme

1 mark for selecting the combination of increasing the concentration gradient and increasing the temperature.
Question 22 · multiple-choice
1 marks
Plant cells with a cytoplasm water potential equivalent to a 0.4 mol/dm3 sucrose solution are placed in different test tubes. In which concentration of sucrose solution will the cells become turgid?
  1. A.0.2 mol/dm3 sucrose solution
  2. B.0.4 mol/dm3 sucrose solution
  3. C.0.6 mol/dm3 sucrose solution
  4. D.0.8 mol/dm3 sucrose solution
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Worked solution

For plant cells to become turgid, water must enter the cells by osmosis. This occurs when cells are placed in a solution with a higher water potential (lower solute concentration) than their cytoplasm. A 0.2 mol/dm3 solution is more dilute than the 0.4 mol/dm3 cytoplasm, so water enters the cells, making them turgid.

Marking scheme

1 mark for identifying the 0.2 mol/dm3 sucrose solution as the one that will make cells turgid.
Question 23 · multiple-choice
1 marks
In a plant species, the allele for red flowers (R) is dominant over the allele for white flowers (r). A heterozygous plant is crossed with a white-flowered plant. What is the expected ratio of phenotypes in the offspring?
  1. A.1 red : 1 white
  2. B.3 red : 1 white
  3. C.All red
  4. D.All white
Show answer & marking scheme

Worked solution

The heterozygous plant has the genotype Rr. The white-flowered plant must be homozygous recessive (rr). Crossing Rr with rr produces offspring with genotypes Rr (red) and rr (white) in a 1:1 ratio. Therefore, the expected phenotypic ratio is 1 red : 1 white.

Marking scheme

1 mark for determining the correct phenotypic ratio of 1 red : 1 white.
Question 24 · multiple-choice
1 marks
Which statement represents an ecological reason for conserving a tropical rainforest ecosystem?
  1. A.To increase the global production of timber for construction purposes
  2. B.To maintain biodiversity and preserve potential sources of new medicines
  3. C.To increase the land area available for intensive agriculture and farming
  4. D.To reduce the overall volume of water cycled through the local environment
Show answer & marking scheme

Worked solution

Conserving tropical rainforests helps maintain high levels of biodiversity, protecting various species of plants, animals, and microorganisms, which can serve as crucial sources for discovering new medicines and maintaining stable food webs.

Marking scheme

1 mark for identifying the maintenance of biodiversity and preservation of potential medicine sources as the correct reason.
Question 25 · multiple-choice
1 marks
A student measures the rate of water uptake of a leafy shoot using a potometer under different conditions. In set-up X, they place a fan blowing air past the shoot. In set-up Y, they cover the plant with a clear plastic bag. In set-up Z, they increase the room temperature from \(20^\circ\text{C}\) to \(30^\circ\text{C}\). Which combination of changes to the environmental conditions will decrease the rate of transpiration compared to the control (still air, \(20^\circ\text{C}\), uncovered)?
  1. A.Set-up X only
  2. B.Set-up Y only
  3. C.Set-up X and Z
  4. D.Set-up Y and Z Holt-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-t-O’t, extit{S. aureus} was not predicted to form high biofilm under any of the growth temperatures with inulin and showed growth rates between $0.057 \times 10^{-2}$ to $0.076 \times 10^{-2}$ $\text{min}^{-1}$ in pre-adapted cells. These findings show that exposure to pre-incubation environments prior to the addition of prebiotics could have a critical impact on the competitive fitness and relative abundance of probiotic bacteria and can potentially enhance their colonisation success when combined in synbiotic formulations. 2.15 Inulin fermentation properties of pure and mixed cultures of probiotic bacteria in the presence of pathogenic $E$. coli and $S$. aureus as influenced by adaptation temperature. In vitro static human gut model system. Inoculation and incubation of media with single cultures of $P$. acidilactici, $L$. acidophilus, $E$. coli and $S$. aureus, and also their co-culture with the target pathogen (pathogen-probiotic and pathogen-pathogen-probiotic mix) under and $37^\text{o}\text{C}$ and $43^\text{o}\text{C}$. Bacterial growth was measured spectrophotometrically at $600\text{ nm}$ at intervals of $0$, $4$, $8$, $12$, $24$ and $48\text{ h}$. The relative abundance and specific growth rate of each bacterial species was determined using selective media plate count method. Pre-incubation at $43^\text{o}\text{C}$ significantly enhanced the growth of $L$. acidophilus in inulin. In the co-cultures, the density of both $P$. acidilactici and $L$. acidophilus increased at $43^\text{o}\text{C}$ in the presence of $E$. coli and $S$. aureus, relative to mono-cultures. High biofilm formation by $L. acidophilus$ was observed only at $43^\text{o}\text{C}$, whereas $P$. acidilactici, $E$. coli, and $S$. aureus were weak biofilm formers. Pre-incubation at $43^\text{o}\text{C}$ significantly enhanced the competitive advantage of $L. acidophilus$ in inulin, but had a negative effect on $P$. acidilactici. In mixed cultures, $L. acidophilus$ was more competitive than $P. acidilactici$ against $E$. coli and $S$. aureus. The specific growth rate of $L. acidophilus$ was higher at $43^\text{o}\text{C}$ compared to $37^\text{o}\text{C}$ in mono-cultures and in co-cultures. In the pre-adapted cells, the specific growth rate of $L. acidophilus$ was $0.187 \times 10^{-2}$ $\text{min}^{-1}$ in mono-culture and $0.198 \times 10^{-2}$ $\text{min}^{-1}$ in co-culture. The specific growth rate of $P. acidilactici$ was $0.093 \times 10^{-2}$ $\text{min}^{-1}$ in mono-culture and $0.089 \times 10^{-2}$ $\text{min}^{-1}$ in co-culture. In contrast, $E. coli$ was predicted to form high biofilm at $37^\text{o}\text{C}$ but not at $43^\text{o}\text{C}$. The specific growth rate of $E. coli$ was $0.124 \times 10^{-2}$ $\text{min}^{-1}$ at $37^\text{o}\text{C}$ and $0.098 \times 10^{-2}$ $\text{min}^{-1}$ at $43^\text{o}\text{C}$. Finally, $S. aureus$ was not predicted to form high biofilm under any of the growth temperatures with inulin and showed growth rates between $0.057 \times 10^{-2}$ to $0.076 \times 10^{-2}$ $\text{min}^{-1}$ in pre-adapted cells. These findings show that exposure to pre-incubation environments prior to the addition of prebiotics could have a critical impact on the competitive fitness and relative abundance of probiotic bacteria and can potentially enhance their colonisation success when combined in synbiotic formulations. 1 1 1.0
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Worked solution

Covering the plant with a clear plastic bag (Set-up Y) traps water vapour released by transpiration, increasing the humidity around the leaves. This reduces the water vapour concentration gradient between the inside of the leaf and the surrounding air, thereby decreasing the rate of transpiration. Using a fan (Set-up X) or increasing the temperature (Set-up Z) increases the rate of transpiration.

Marking scheme

1 mark for identifying that increasing humidity (Set-up Y) decreases transpiration while fan (Set-up X) and higher temperature (Set-up Z) increase it, leading to correct option B.
Question 26 · multiple_choice
1 marks
A student uses a potometer to measure the rate of water uptake of a leafy shoot. Which combination of environmental factors will result in the shortest time for the bubble to move a distance of 10 cm?
  1. A.High humidity, high temperature, and still air
  2. B.High humidity, low temperature, and moving air
  3. C.Low humidity, high temperature, and moving air
  4. D.Low humidity, low temperature, and still air
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Worked solution

Transpiration is the loss of water vapour from plant leaves. The rate of transpiration is increased by low humidity, high temperature, and moving air (wind). A higher rate of transpiration increases the rate of water uptake, causing the bubble in the potometer to move faster, thus taking the shortest time to cover the specified distance.

Marking scheme

Award 1 mark for identifying the correct combination (low humidity, high temperature, and moving air) which maximizes transpiration rate, resulting in the fastest bubble movement.
Question 27 · multiple_choice
1 marks
The human heart consists of four muscular chambers. Which chamber contracts to pump oxygenated blood at the highest pressure to the rest of the body, and what structural feature allows this?
  1. A.Left atrium, which has a thin muscular wall
  2. B.Left ventricle, which has a thick muscular wall
  3. C.Right atrium, which has a thick muscular wall
  4. D.Right ventricle, which has a thin muscular wall
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Worked solution

The left ventricle is responsible for pumping oxygenated blood through the aorta to the systemic circulation (all body tissues except the lungs). Because it must push blood over a long distance against high resistance, it requires a very thick, muscular wall to generate the necessary high pressure.

Marking scheme

Award 1 mark for identifying the left ventricle and its thick muscular wall as the correct chamber and structural adaptation.
Question 28 · multiple_choice
1 marks
During ultrafiltration in the human kidney, which substances from the blood pass into the Bowman's capsule, and which substance is fully reabsorbed back into the blood at the proximal convoluted tubule under normal conditions?
  1. A.Filtered: red blood cells and glucose; Reabsorbed: glucose
  2. B.Filtered: water, salts, glucose, and urea; Reabsorbed: glucose
  3. C.Filtered: water, proteins, and salts; Reabsorbed: proteins
  4. D.Filtered: urea and plasma proteins; Reabsorbed: urea
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Worked solution

Ultrafiltration allows small molecules like water, salts, glucose, and urea to pass from the glomerulus into the Bowman's capsule, while blood cells and large proteins remain in the capillary. In the proximal convoluted tubule, glucose is actively and fully reabsorbed back into the blood.

Marking scheme

Award 1 mark for selecting the option that correctly identifies the components of the filtrate (water, salts, glucose, and urea) and the substance fully reabsorbed (glucose).
Question 29 · multiple_choice
1 marks
A student tests a liquid food sample using three reagents. The Benedict's test (after heating) gives a brick-red precipitate, the Biuret test turns violet, and the iodine test remains orange-brown. Which biological molecules are present in this food sample?
  1. A.Reducing sugar and starch only
  2. B.Protein and starch only
  3. C.Reducing sugar and protein only
  4. D.Reducing sugar, protein, and starch
Show answer & marking scheme

Worked solution

A positive Benedict's test (brick-red) indicates the presence of reducing sugar. A positive Biuret test (violet) indicates the presence of protein. A negative iodine test (orange-brown remaining, not blue-black) indicates that starch is absent. Therefore, only reducing sugar and protein are present.

Marking scheme

Award 1 mark for correctly matching the positive Benedict's and Biuret tests with the presence of reducing sugar and protein, and the negative iodine test with the absence of starch.
Question 30 · multiple_choice
1 marks
Under which set of conditions will the rate of diffusion of gas molecules across a membrane be the highest?
  1. A.High temperature, short diffusion distance, and a steep concentration gradient
  2. B.High temperature, long diffusion distance, and a shallow concentration gradient
  3. C.Low temperature, short diffusion distance, and a shallow concentration gradient
  4. D.Low temperature, long diffusion distance, and a steep concentration gradient
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Worked solution

Diffusion rate is increased by a higher temperature (increases kinetic energy of molecules), a shorter diffusion distance (less distance to travel), and a steeper concentration gradient (greater difference in concentration).

Marking scheme

Award 1 mark for identifying the combination of factors (high temperature, short distance, steep gradient) that maximizes the rate of diffusion.
Question 31 · multiple_choice
1 marks
An epidermal cell from a red onion is placed in a highly concentrated sucrose solution. What is the net direction of water movement and the resulting state of the cell?
  1. A.Water moves into the cell by osmosis; the cell becomes turgid
  2. B.Water moves into the cell by active transport; the cell lyses
  3. C.Water moves out of the cell by osmosis; the cell becomes plasmolysed
  4. D.Water moves out of the cell by diffusion; the cell becomes turgid
Show answer & marking scheme

Worked solution

A concentrated sucrose solution has a lower water potential than the cytoplasm of the onion cell. Therefore, water moves out of the cell down a water potential gradient by osmosis. This causes the vacuole and cytoplasm to shrink, pulling the cell membrane away from the cell wall, leaving the cell plasmolysed.

Marking scheme

Award 1 mark for identifying that water moves out of the cell by osmosis and the cell becomes plasmolysed.
Question 32 · multiple_choice
1 marks
In a type of plant, the allele for red flowers (R) is dominant to the allele for white flowers (r). Two heterozygous plants are crossed. What is the probability that an offspring will have white flowers?
  1. A.0.00 (0%)
  2. B.0.25 (25%)
  3. C.0.50 (50%)
  4. D.0.75 (75%)
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Worked solution

Crossing two heterozygous plants (Rr x Rr) results in offspring genotypes in a ratio of 1 RR : 2 Rr : 1 rr. Only the homozygous recessive genotype (rr) produces white flowers. This represents a probability of 1/4, which is 25%.

Marking scheme

Award 1 mark for calculating the probability of the homozygous recessive phenotype as 25% using a monohybrid cross.
Question 33 · multiple_choice
1 marks
Which conservation strategy provides the most sustainable balance between protecting endangered wildlife and supporting the local human population?
  1. A.Completely banning humans from entering the entire natural habitat
  2. B.Introducing non-native predators to manage native herbivore populations
  3. C.Establishing protected areas with regulated ecotourism and sustainable resource harvesting
  4. D.Removing all remaining individuals to breed them exclusively in captivity in other countries
Show answer & marking scheme

Worked solution

Sustainable conservation involves protecting ecosystems while allowing human activities that do not deplete resources. Regulated ecotourism and sustainable harvesting of materials provide economic benefits to local communities, ensuring their long-term support for conservation efforts.

Marking scheme

Award 1 mark for selecting the option that balances ecosystem protection with sustainable human resource use and ecotourism.

Paper 43

Answer all structured questions on the question paper.
7 Question · 80.00999999999999 marks
Question 1 · Structured
11.43 marks
(a) Define the term transpiration. [2] (b) Describe how a student would use a bubble potometer to estimate the rate of water uptake in a leafy shoot. Include one precaution that must be taken when setting up the apparatus. [4] (c) Explain, in terms of water potential gradient and kinetic energy of water molecules, how increasing the wind speed affects the rate of transpiration. [3] (d) State two anatomical adaptations of xerophytic leaves that reduce water loss. [2]
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Worked solution

(a) Transpiration is the loss of water vapour from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by diffusion of water vapour through the stomata. (b) To use a potometer: cut the leafy shoot under water to prevent air entering xylem vessels; insert the shoot into the potometer tube and seal all connections with petroleum jelly to make it airtight; introduce a single air bubble into the capillary tube; measure the distance moved by the bubble in a set time period. (c) Increasing wind speed blows away the water vapour accumulating near the leaf surface. This maintains a steep water potential gradient between the leaf interior and the outside air. Consequently, diffusion of water vapour out of the stomata occurs at a faster rate. (d) Two adaptations: thick waxy cuticle, sunken stomata, rolled leaves, or leaf hairs.

Marking scheme

(a) [Max 2 marks]: loss of water vapour [1], from leaves/evaporation from mesophyll surfaces [1], followed by diffusion through stomata [1]. (b) [Max 4 marks]: cut shoot underwater [1], seal joints with petroleum jelly/make airtight [1], introduce air bubble [1], measure distance bubble moves in a set time [1]. (c) [Max 3 marks]: wind removes water vapour from leaf surface [1], maintains steep water potential gradient [1], increases rate of diffusion [1]. (d) [Max 2 marks]: award 1 mark for any two valid adaptations: thick waxy cuticle [1], sunken stomata [1], rolled leaves [1], leaves reduced to spines [1], hairs on leaves [1].
Question 2 · Structured
11.43 marks
(a) Name the blood vessels that supply the muscle tissue of the heart with oxygen and nutrients, and explain the consequences of these vessels becoming blocked. [4] (b) The muscular wall of the left ventricle is significantly thicker than that of the right ventricle. Explain the physiological importance of this structural difference. [3] (c) Describe the function of the septum in the mammalian heart. [2] (d) State two lifestyle factors, other than diet, that increase the risk of developing coronary heart disease (CHD). [2]
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Worked solution

(a) The vessels are the coronary arteries. Blockage of these vessels restricts the supply of oxygen and glucose to cardiac muscle cells, preventing aerobic respiration. The cells must respire anaerobically, producing lactic acid which is toxic, eventually causing cardiac muscle cells to die and leading to a heart attack. (b) The left ventricle must pump blood under high pressure to the entire systemic circulation (all body organs), which has high resistance. The right ventricle only pumps blood to the lungs (pulmonary circulation), which is nearby and requires lower pressure to prevent damage to delicate lung capillaries. (c) The septum physically separates oxygenated blood on the left side from deoxygenated blood on the right side, preventing mixing and ensuring highly oxygenated blood is sent to the body tissues. (d) Risk factors include: smoking, chronic stress, lack of physical exercise, or genetic history.

Marking scheme

(a) [Max 4 marks]: coronary arteries [1], blocks oxygen/glucose supply [1], prevents aerobic respiration/forces anaerobic respiration [1], cardiac muscle cells die/heart attack occurs [1]. (b) [Max 3 marks]: left ventricle pumps blood to body/systemic circulation [1], needs high pressure/overcomes high resistance [1], right ventricle pumps to lungs/pulmonary circulation at lower pressure to prevent capillary damage [1]. (c) [Max 2 marks]: separates oxygenated and deoxygenated blood [1], prevents mixing of blood to maintain high oxygen concentration [1]. (d) [Max 2 marks]: any two from: smoking [1], lack of exercise [1], stress [1], genetic predisposition [1].
Question 3 · Structured
11.43 marks
(a) Define excretion and name two toxic waste substances, excluding carbon dioxide, excreted by the human kidney. [3] (b) Describe the process of ultrafiltration in the renal corpuscle (glomerulus and Bowman's capsule). [4] (c) State the exact region of the nephron where glucose is selectively reabsorbed and the transport process used. [2] (d) Dehydration leads to the production of concentrated urine. Explain how the body adjusts urine concentration in response to low water potential of the blood. [2]
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Worked solution

(a) Excretion is the removal from organisms of toxic materials, the waste products of metabolism, and substances in excess of requirements. The two toxic waste substances excreted by the kidney are urea and excess mineral salts. (b) High blood pressure is created in the glomerulus because the afferent arteriole is wider than the efferent arteriole. This pressure forces small molecules (water, glucose, amino acids, urea, and ions) through the basement membrane (acting as a filter) into the Bowman's capsule to form glomerular filtrate. Large proteins and blood cells are too big to pass and remain in the blood. (c) Glucose is selectively reabsorbed in the proximal convoluted tubule by active transport. (d) The brain (hypothalamus/pituitary) detects low water potential and secretes more ADH (antidiuretic hormone). ADH increases the permeability of the collecting ducts, allowing more water to be reabsorbed back into the blood by osmosis, resulting in a low volume of concentrated urine.

Marking scheme

(a) [Max 3 marks]: definition of excretion [1], urea [1], excess salts/ions [1]. (b) [Max 4 marks]: high blood pressure in glomerulus [1], small molecules forced out of capillaries [1], basement membrane acts as a filter [1], large proteins/blood cells remain in blood [1], filtrate enters Bowman's capsule [1]. (c) [Max 2 marks]: proximal convoluted tubule [1], active transport [1]. (d) [Max 2 marks]: pituitary releases ADH [1], collecting duct becomes more permeable to water/more water reabsorbed into blood [1].
Question 4 · Structured
11.43 marks
(a) Describe how you would test a liquid food sample to demonstrate the presence of: (i) reducing sugars, and (ii) proteins. Include the names of the reagents used and the positive color results. [4] (b) State the four main chemical elements present in all proteins and identify the specific element present in proteins but absent in lipids and carbohydrates. [3] (c) Explain how the unique sequence of amino acids determines the function of an enzyme. [2] (d) Describe the structural shape of a DNA molecule and name the complementary base pairs. [2]
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Worked solution

(a) (i) To test for reducing sugars: Add Benedict's solution to the sample and heat in a hot water bath. A positive result changes from blue to green, yellow, or brick-red. (ii) To test for proteins: Add Biuret solution to the sample. A positive result changes from blue to purple or violet. (b) The four elements in all proteins are Carbon, Hydrogen, Oxygen, and Nitrogen. Nitrogen is present in proteins but absent in simple lipids and carbohydrates. (c) The specific sequence of amino acids determines how the protein chain folds into a precise three-dimensional shape. This shape determines the specific structure of the active site, which must be complementary to the substrate molecule for catalysis. (d) DNA is a double helix composed of two strands joined by cross-links of bases. The complementary base pairs are Adenine (A) with Thymine (T), and Cytosine (C) with Guanine (G).

Marking scheme

(a) [Max 4 marks]: Benedict's test + heat [1], reducing sugar positive is green/yellow/brick-red [1], Biuret test [1], protein positive is purple/violet [1]. (b) [Max 3 marks]: Carbon, Hydrogen, Oxygen, Nitrogen (all 4 listed) [2], (award 1 mark if only 2-3 listed), Nitrogen is absent in carbohydrates/lipids [1]. (c) [Max 2 marks]: sequence determines 3D folding/shape [1], determines the shape of the active site/complementary to substrate [1]. (d) [Max 2 marks]: double helix [1], A-T and C-G complementary base pairing [1].
Question 5 · Structured
11.43 marks
(a) Define the term osmosis. [3] (b) A group of students placed potato cylinders into a concentrated sucrose solution. After 2 hours, they observed a decrease in the mass of the cylinders. Explain this observation in terms of water potential. [4] (c) Describe the appearance of a plant cell that has lost a large volume of water by osmosis and state the term used to describe this condition. [2] (d) Explain why a human red blood cell lyses (bursts) when placed in distilled water, whereas a plant cell does not. [2]
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Worked solution

(a) Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane. (b) The concentrated sucrose solution has a lower water potential than the cytoplasm/vacuole of the potato cells. Water moves out of the potato cells by osmosis, down a water potential gradient, through the partially permeable cell membranes. This loss of water decreases the volume and mass of the potato tissue. (c) The vacuole shrinks and the cell membrane pulls away from the rigid cell wall. This state is called plasmolysis (the cell is plasmolysed). (d) Red blood cells lack a cell wall, so when water enters by osmosis, they swell until the membrane ruptures. Plant cells have a strong cellulose cell wall that resists internal pressure, preventing the cell from bursting and making it turgid.

Marking scheme

(a) [Max 3 marks]: net movement of water molecules [1], higher to lower water potential [1], partially permeable membrane [1]. (b) [Max 4 marks]: sucrose solution has lower water potential than potato cells [1], water potential gradient established [1], water moves out of cells [1], by osmosis/across partially permeable membrane [1], reduction in cell volume/mass [1]. (c) [Max 2 marks]: cytoplasm/vacuole shrinks or membrane pulls away from cell wall [1], plasmolysis/plasmolysed [1]. (d) [Max 2 marks]: red blood cell has no cell wall so pressure causes rupture [1], plant cell has rigid cellulose cell wall that prevents bursting/reaches turgidity [1].
Question 6 · Structured
11.43 marks
(a) Define the term endangered species and list two human activities that can cause a species to become endangered. [3] (b) Explain how the implementation of (i) fishing quotas and (ii) large-mesh size nets can help to conserve wild fish populations. [4] (c) Explain why it is important to conserve ecosystems and maintain biodiversity. [4]
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Worked solution

(a) An endangered species is a species that is at risk of extinction. Human activities that cause endangerment include habitat destruction (deforestation), pollution, climate change, overhunting, and introduction of invasive species. (b) (i) Fishing quotas limit the mass or number of fish that can be legally caught, preventing overfishing and ensuring a sustainable breeding population remains. (ii) Large-mesh nets allow smaller, juvenile fish to pass through unharmed, ensuring they survive to reach sexual maturity and reproduce. (c) Ecosystem conservation and biodiversity are vital because: they maintain food web stability (preventing collapse of trophic levels); they preserve potential future resources, such as medicines or genes for crop breeding; they maintain vital ecosystem services like nutrient cycling, oxygen production, and soil protection; and they provide aesthetic, educational, and ecotourism value.

Marking scheme

(a) [Max 3 marks]: definition: species at risk of extinction [1], two activities: deforestation/habitat destruction [1], overhunting/overfishing [1], pollution [1]. (b) [Max 4 marks]: (i) quotas limit catch [1], ensures sustainable breeding population survives [1], (ii) large-mesh nets let young fish escape [1], young fish grow to maturity and reproduce [1]. (c) [Max 4 marks]: food web stability/prevents trophic collapse [1], source of medicines/genetic resources [1], ecosystem services (oxygen, cycling, erosion control) [1], ethical/ecotourism values [1].
Question 7 · Structured
11.43 marks
(a) Define the term enzyme and explain the lock and key hypothesis of enzyme action. [4] (b) Describe and explain the effect of: (i) a low temperature of 5 degrees C and (ii) an extremely high temperature of 70 degrees C on the rate of an enzyme-controlled reaction. [4] (c) State three variables that must be controlled when investigating the effect of pH on the activity of salivary amylase. [3]
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Worked solution

(a) An enzyme is a protein that functions as a biological catalyst. According to the lock and key hypothesis, the enzyme (the lock) has an active site with a specific 3D shape that is complementary to the shape of the substrate (the key). The substrate binds to the active site to form an enzyme-substrate complex, where chemical bonds are broken/made, releasing products. (b) (i) At 5 degrees C, molecules have low kinetic energy and move slowly, resulting in few collisions per second between enzyme and substrate and a low reaction rate. (ii) At 70 degrees C, excess thermal energy breaks the hydrogen bonds maintaining the enzyme's 3D structure. The shape of the active site is permanently altered (denatured), so the substrate can no longer fit, and the reaction stops. (c) Three variables: temperature, concentration of starch (substrate), concentration of amylase (enzyme), volume of starch/amylase solutions, and incubation duration.

Marking scheme

(a) [Max 4 marks]: protein [1], biological catalyst [1], specific 3D active site complementary to substrate [1], enzyme-substrate complex forms to yield products [1]. (b) [Max 4 marks]: low temperature: low kinetic energy [1], fewer successful collisions per unit time [1], high temperature: denatures enzyme [1], active site shape changed/substrate can no longer bind [1]. (c) [Max 3 marks]: award 1 mark for each valid variable: temperature [1], amylase concentration [1], starch concentration [1], volume of solutions [1].

Paper 63

Answer all practical-style alternative questions.
2 Question · 40 marks
Question 1 · Practical Alternative
20 marks
An investigation was carried out to study the effect of wind speed on the rate of transpiration from a leafy shoot. A student set up a balance with a conical flask containing water and a leafy shoot. A layer of oil was placed on top of the water to prevent evaporation. A fan was set up at different distances from the shoot to vary the wind speed: 10 cm, 30 cm, and 50 cm. A control experiment was also set up with no fan. The mass of the flask was measured at the start and after exactly 2 hours. The results are as follows: (i) No fan: Start mass = 152.4 g, End mass = 150.8 g. (ii) 50 cm distance: Start mass = 151.8 g, End mass = 148.6 g. (iii) 30 cm distance: Start mass = 153.1 g, End mass = 148.3 g. (iv) 10 cm distance: Start mass = 150.5 g, End mass = 143.7 g. Tasks: 1. Calculate the mass loss (g) for each condition, and then calculate the rate of water loss in grams per hour (g/h). [4 marks] 2. Identify the independent variable and the dependent variable in this investigation. [2 marks] 3. Explain why a layer of oil was added to the surface of the water in each flask. [2 marks] 4. Describe how you would plot a line graph of these results, stating what should be on each axis and the trend you would expect to see for the distances 10 cm, 30 cm, and 50 cm. [4 marks] 5. Identify two variables (other than wind speed) that must be kept constant (controlled variables) during this investigation, and describe how they could be controlled. [4 marks] 6. Describe a chemical test the student could use to confirm that the liquid lost by the plant is water, including the expected positive result. [4 marks]
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Worked solution

1. To find the mass loss: Subtract the End mass from the Start mass. No fan: 152.4 - 150.8 = 1.6 g. 50 cm: 151.8 - 148.6 = 3.2 g. 30 cm: 153.1 - 148.3 = 4.8 g. 10 cm: 150.5 - 143.7 = 6.8 g. To find the rate of water loss per hour: Divide the mass loss by 2 hours. No fan: 1.6 / 2 = 0.8 g/h. 50 cm: 3.2 / 2 = 1.6 g/h. 30 cm: 4.8 / 2 = 2.4 g/h. 10 cm: 6.8 / 2 = 3.4 g/h. 2. The independent variable is the factor changed by the investigator (distance of fan / wind speed). The dependent variable is the factor measured (mass loss or rate of water loss). 3. Oil forms an impermeable barrier on the water surface, stopping water molecules from escaping directly into the air via evaporation. 4. Graphs must have labeled axes with units. The independent variable goes on the horizontal (x) axis, and the dependent variable on the vertical (y) axis. The trend shows that the rate of transpiration is higher when the fan is closer because fast-moving air removes water vapor from near the leaves, maintaining a steep concentration gradient. 5. Controlled variables must be kept the same to make the investigation valid. Temperature affects kinetic energy and humidity affects the concentration gradient. 6. Pure water chemically reacts with anhydrous copper(II) sulfate to hydrate it, producing a characteristic color change from white to blue.

Marking scheme

1. [4 marks total]: 1 mark for calculating all correct mass losses (1.6 g, 3.2 g, 4.8 g, 6.8 g). 1 mark for correct division by 2 to find rate. 2 marks for all four correct rates (0.8, 1.6, 2.4, 3.4) with correct units (g/h). 2. [2 marks total]: 1 mark for identifying independent variable (wind speed / distance of fan). 1 mark for identifying dependent variable (mass loss / rate of water loss). 3. [2 marks total]: 1 mark for stating it prevents evaporation from the flask water surface. 1 mark for explaining that this ensures mass loss is only due to transpiration/water uptake by the plant. 4. [4 marks total]: 1 mark for correct axis labels with units: x-axis = Distance of fan / cm, y-axis = Rate of water loss / (g/h). 1 mark for linear, appropriate scales using at least half of the grid. 1 mark for plotting all three distance points accurately. 1 mark for describing the trend: rate of water loss decreases as the distance of the fan increases. 5. [4 marks total]: 2 marks for identifying two valid controlled variables (e.g., temperature, light intensity, humidity, size/surface area of leafy shoot). 2 marks for describing how each identified variable is controlled (e.g., use a thermometer/air conditioner, keep at a fixed distance from a lamp, use the same shoot). 6. [4 marks total]: 1 mark for naming a suitable reagent (anhydrous copper(II) sulfate or anhydrous cobalt chloride paper). 1 mark for initial state/color of the reagent (white for copper sulfate, blue for cobalt chloride paper). 1 mark for adding the liquid to the reagent. 1 mark for the correct positive color change (turns blue for copper sulfate, turns pink for cobalt chloride paper).
Question 2 · Practical Alternative
20 marks
Catalase is an enzyme found in plant tissues, such as potato, which breaks down hydrogen peroxide to form water and oxygen gas. Part 1: A student drew a diagram of a potato cell observed under a light microscope. The length of the cell in the drawing is 62 mm. The actual length of the cell is 0.08 mm. 1. Calculate the magnification of the student's drawing. Show your working and state the answer to the nearest whole number. [3 marks] 2. State two rules of biological drawing that the student should follow to ensure their diagram is scientific. [2 marks] Part 2: Plan an investigation to determine the effect of temperature on the rate of oxygen production by potato catalase. Include details of the apparatus, how you will vary the temperature, which variables you will control, how you will measure the dependent variable, and how you will ensure the results are reliable. [6 marks] Part 3: The student carried out a preliminary experiment to compare the catalase activity in raw potato and cooked (boiled) potato. They measured the volume of oxygen gas produced in 3 minutes. Raw potato trials: 12.4 cm3, 13.0 cm3, 12.1 cm3. Cooked potato trials: 0.0 cm3, 0.0 cm3, 0.0 cm3. 3. Calculate the mean volume of gas produced by the raw potato. Show your working and round your answer to one decimal place. [2 marks] 4. Explain the difference in the volume of oxygen gas produced between raw potato and cooked potato. [3 marks] 5. Describe a biochemical test the student could use to confirm that the potato tissue contains starch, including the positive result. [2 marks] 6. State one safety hazard associated with using hydrogen peroxide and suggest an appropriate precaution. [2 marks]
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Worked solution

1. To calculate magnification, ensure both values are in the same units. Drawing length = 62 mm. Actual length = 0.08 mm. Formula: Magnification (M) = Image size (I) / Actual size (A). Calculation: M = 62 / 0.08 = 775. State with a 'x' symbol: x775. 2. Scientific biological drawings must use clean single lines (no sketching), have no shading/coloring, and accurately depict proportions. 3. Mean volume calculation: Sum of trials = 12.4 + 13.0 + 12.1 = 37.5 cm3. Divide by the number of trials: 37.5 / 3 = 12.5 cm3. 4. Heating enzymes above their optimum temperature breaks bonds holding their tertiary structure together, denaturing them. Since the active site changes shape, no catalytic activity can take place in boiled potato. 5. Starch is detected using iodine dissolved in potassium iodide solution. A positive reaction yields a distinct dark blue-black complex. 6. Hydrogen peroxide is a strong oxidizing agent and chemical irritant; personal protective equipment (PPE) such as safety spectacles must be worn.

Marking scheme

1. [3 marks total]: 1 mark for stating the correct formula (Magnification = Image / Actual). 1 mark for correct substitution (62 / 0.08). 1 mark for correct final answer of 775 or x775 (accept 775 times, reject decimal values like 775.0). 2. [2 marks total]: Accept any two of: clear single lines/no sketching, no shading/coloring, large size (uses over half the space), includes clear labeling lines drawn with a ruler. 3. [2 marks total]: 1 mark for correct working (summing values to get 37.5 and dividing by 3). 1 mark for correct final answer of 12.5 cm3. 4. [3 marks total]: 1 mark for stating that boiling denatures the catalase enzyme. 1 mark for explaining that the active site changes shape. 1 mark for stating that the substrate (hydrogen peroxide) can no longer fit/bind (no enzyme-substrate complexes can form). 5. [2 marks total]: 1 mark for adding iodine solution/iodine. 1 mark for stating the color changes from orange/brown to blue-black. 6. [2 marks total]: 1 mark for identifying the hazard: hydrogen peroxide is an irritant/corrosive/can damage eyes or skin. 1 mark for a matching precaution: wear safety goggles/gloves.

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