An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 OCR GCSE (9-1) Gateway Science - Biology A - J247 paper. Not affiliated with or reproduced from OCR.
Section A
Answer all questions. You should spend a maximum of 30 minutes on this section.
15 Question · 15 marks
Question 1 · Multiple Choice
1 marks
A model of a cell is represented by a cube. Cell X has a side length of \(2\text{ cm}\). Cell Y has a side length of \(6\text{ cm}\). What is the surface area to volume ratio (SA:V) of Cell X and Cell Y, and how does it change as cell size increases?
A.Cell X is \(3:1\), Cell Y is \(1:1\); the ratio decreases as size increases.
B.Cell X is \(1:3\), Cell Y is \(1:1\); the ratio increases as size increases.
C.Cell X is \(6:1\), Cell Y is \(2:1\); the ratio decreases as size increases.
D.Cell X is \(3:1\), Cell Y is \(3:1\); the ratio remains constant as size increases.
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Worked solution
For Cell X with a side length of \(2\text{ cm}\): Surface Area = \(6 \times (2^2) = 24\text{ cm}^2\), Volume = \(2^3 = 8\text{ cm}^3\). The SA:V ratio is \(24 : 8 = 3:1\). For Cell Y with a side length of \(6\text{ cm}\): Surface Area = \(6 \times (6^2) = 216\text{ cm}^2\), Volume = \(6^3 = 216\text{ cm}^3\). The SA:V ratio is \(216 : 216 = 1:1\). Therefore, as size increases, the SA:V ratio decreases from \(3:1\) to \(1:1\).
Marking scheme
1 mark for the correct answer A. (AO2)
Question 2 · Multiple Choice
1 marks
Which sequence correctly shows the pathway of an electrical impulse along a reflex arc when a person withdraws their hand from a hot object?
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Worked solution
A reflex arc starts with a receptor detecting a stimulus, sending an electrical impulse along a sensory neuron to the central nervous system (spinal cord). The impulse is passed across a synapse to a relay neuron, and then across another synapse to a motor neuron, which carries it to the effector (such as a muscle) to bring about a response.
Marking scheme
1 mark for the correct answer A. (AO1)
Question 3 · Multiple Choice
1 marks
An experiment is carried out to investigate the rate of photosynthesis in Elodea at \(20^\circ\text{C}\). At high light intensities, the rate of photosynthesis is found to be much higher when the water contains \(0.15\%\) sodium hydrogencarbonate (a source of carbon dioxide) compared to \(0.04\%\). What is the limiting factor at high light intensity when the carbon dioxide concentration is \(0.04\%\)?
A.Carbon dioxide concentration
B.Light intensity
C.Temperature
D.Water availability
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Worked solution
At high light intensity, light is no longer the limiting factor. Because increasing the concentration of carbon dioxide from \(0.04\%\) to \(0.15\%\) increases the rate of photosynthesis, carbon dioxide concentration must have been the limiting factor at the lower concentration.
Marking scheme
1 mark for the correct answer A. (AO2)
Question 4 · multiple_choice
1 marks
An experiment was carried out to investigate the effect of temperature on the rate of an enzyme-controlled reaction. Which statement correctly explains why the reaction rate decreases at temperatures higher than the optimum temperature?
A.The reactant molecules have less kinetic energy, leading to fewer successful collisions.
B.The shape of the enzyme's active site changes, so the substrate can no longer fit.
C.The enzyme molecules are completely broken down into individual amino acids.
D.The substrate molecules denature and can no longer bind to the active site of the enzyme27s molecule32s structure33s state34s shape35s active site36s form37s shape38s form39s active site40s active site41s active site42s structure43s active site44s active site45s active site46s active site47s active site48s active site49s active site50s active site51s active site52s active site53s active site54s active site55s active site56s active site57s active site58s active site59s active site60s active site61s active site62s active site63s active site64s active site65s active site66s active site67s active site68s active site69s active site70s active site71s active site72s active site73s active site74s active site75s active site76s active site77s active site78s active site79s active site80s active site81s active site82s active site83s active site84s active site85s active site86s active site87s active site88s active site89s active site90s active site91s active site92s active site93s active site94s 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Worked solution
At temperatures above the optimum, the increased thermal energy causes the enzyme molecule to vibrate excessively. This breaks the weak bonds (such as hydrogen bonds) that maintain the specific three-dimensional shape of the enzyme. As a result, the shape of the active site changes permanently (denaturation), so the substrate molecule is no longer complementary and cannot bind to form an enzyme-substrate complex.
Marking scheme
[1 mark] b is correct. Option a describes what happens at very low temperatures. Option c is incorrect because denaturation does not break the peptide bonds to release individual amino acids. Option d is incorrect because it is the enzyme that denatures, not the substrate.
Question 5 · multiple_choice
1 marks
A person has not eaten for several hours. Which of the following describes the correct response of the endocrine system to maintain blood glucose homeostasis?
A.The pancreas secretes glucagon, which stimulates the liver to convert glycogen into glucose.
B.The pancreas secretes insulin, which stimulates body cells to take up more glucose.
C.The adrenal gland secretes adrenaline, which stimulates the liver to convert glucose into glycogen.
D.The pancreas secretes glucagon, which stimulates the liver to convert glucose into glycogen-type storage molecules during low blood glucose conditions.
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Worked solution
When blood glucose levels are low, the pancreas detects this change and secretes the hormone glucagon into the blood. Glucagon travels to the liver, where it triggers the conversion of stored glycogen into glucose, which is then released into the blood to raise the blood glucose concentration back to normal.
Marking scheme
[1 mark] a is correct. Option b describes the response to high blood glucose. Option c is incorrect because adrenaline triggers glycogen breakdown, not glycogen synthesis. Option d is incorrect because glucagon causes glycogen to break down into glucose, not glucose to convert into glycogen.
Question 6 · multiple_choice
1 marks
Doctors are concerned about the rise of antibiotic-resistant bacteria. Which of the following describes the correct sequence of events in the development of antibiotic resistance by natural selection?
A.Mutation occurs -> exposure to antibiotic -> resistant bacteria die -> non-resistant bacteria reproduce
B.Exposure to antibiotic -> mutation occurs to resist it -> resistant bacteria survive and reproduce
C.Mutation occurs -> exposure to antibiotic -> resistant bacteria survive and reproduce -> resistance gene passed to offspring
D.Exposure to antibiotic -> non-resistant bacteria mutate to protect themselves -> all bacteria survive and reproduce
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Worked solution
Natural selection begins with genetic variation within a bacterial population, caused by random mutation. When an antibiotic is introduced, it acts as a selective pressure. The resistant bacteria survive the treatment and reproduce, passing on the allele for resistance to their offspring. This increases the proportion of resistant bacteria over time.
Marking scheme
[1 mark] c is correct. Option a is incorrect because non-resistant bacteria are killed by the antibiotic, not resistant ones. Options b and d are incorrect because they suggest that mutations are caused/directed by the exposure to antibiotics (Lamarckian style), whereas mutations are random events that occur prior to selection pressure.
Question 7 · Multiple Choice
1 marks
A student investigates the rate of photosynthesis in Elodea (pondweed) at different light intensities. The temperature is maintained at any constant temperature such as
(18^\circ\text{C}\) and the carbon dioxide concentration is maintained at \(0.03\%\). At very high light intensities, the rate of photosynthesis remains constant. Which of the following changes would increase the rate of photosynthesis at these very high light intensities?
A.Increasing the carbon dioxide concentration to \(0.1\%\)
B.Decreasing the temperature of the water to \(10^\circ\text{C}\)
C.Increasing the distance of the light source from the pondweed
D.Placing the pondweed in a darker room
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Worked solution
At very high light intensities, light is no longer the limiting factor. The rate is limited by either temperature or carbon dioxide concentration. Increasing the carbon dioxide concentration to \(0.1\%\) removes this limitation, allowing the rate of photosynthesis to increase. Decreasing the temperature would slow down enzyme-controlled reactions, reducing the rate. Moving the light source further away or placing the pondweed in the dark would decrease light intensity, which would decrease or stop photosynthesis.
Marking scheme
1 mark for the correct answer A. Reject all other options.
Question 8 · Multiple Choice
1 marks
Thyroxine is a hormone produced by the thyroid gland that helps regulate metabolic rate. Its release is controlled by negative feedback. Which of the following sequences of events occurs when thyroxine levels in the blood decrease below normal levels?
A.The pituitary gland secretes more TSH, stimulating the thyroid gland to secrete more thyroxine.
B.The pituitary gland secretes less TSH, stimulating the thyroid gland to secrete more thyroxine.
C.The thyroid gland secretes more TSH, stimulating the pituitary gland to secrete more thyroxine.
D.The pituitary gland secretes more thyroxine, stimulating the thyroid gland to secrete more TSH.
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Worked solution
When thyroxine levels in the blood fall below normal, the pituitary gland detects this change and secretes more Thyroid Stimulating Hormone (TSH). TSH stimulates the thyroid gland to release more thyroxine into the blood, restoring the level back to normal. This is a key example of a negative feedback mechanism.
Marking scheme
1 mark for the correct answer A. Reject all other options.
Question 9 · Multiple Choice
1 marks
During an investigation into catalase activity, a student measures that \(15.0\text{ cm}^3\) of oxygen gas is produced over a period of \(2.5\text{ minutes}\). What is the mean rate of reaction in \(cm^3\ s^{-1}\)?
A.0.10
B.6.0
C.10.0
D.0.17
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Worked solution
To calculate the mean rate of reaction in \(\text{cm}^3\text{ s}^{-1}\): 1. Convert the time from minutes to seconds: \(2.5\text{ minutes} \times 60\text{ seconds/minute} = 150\text{ seconds}\). 2. Divide the volume of gas produced by the time taken in seconds: \(\text{Rate} = \frac{15.0\text{ cm}^3}{150\text{ s}} = 0.10\text{ cm}^3\text{ s}^{-1}\).
Marking scheme
1 mark for the correct calculation and answer A. Reject all other options.
Question 10 · multiple_choice
1 marks
An image of a mitochondrion has a length of \( 48\text{ mm} \). The actual length of the mitochondrion is \( 3\text{ }\mu\text{m} \). What is the magnification of the image?
A.\( \times 16 \)
B.\( \times 1,600 \)
C.\( \times 16,000 \)
D.\( \times 160,000 \)
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Worked solution
To calculate magnification, use the formula: \( \text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} \). First, convert the measurements into the same units. Converting millimetres to micrometres: \( 48\text{ mm} = 48,000\text{ }\mu\text{m} \). Next, divide the image size by the actual size: \( \text{Magnification} = \frac{48,000\text{ }\mu\text{m}}{3\text{ }\mu\text{m}} = 16,000 \). Therefore, the correct magnification is \( \times 16,000 \).
Marking scheme
1 mark for the correct answer C. Accept C. Reject all other responses.
Question 11 · multiple_choice
1 marks
Which statement correctly describes how a nerve impulse is transmitted across a synapse?
A.Chemical neurotransmitters diffuse across the synaptic cleft from the post-synaptic neurone to the pre-synaptic neurone.
B.Electrical impulses jump directly across the synaptic cleft via specialized protein channels.
C.Chemical neurotransmitters diffuse across the synaptic cleft and bind to specific receptors on the post-synaptic membrane.
D.Electrical impulses are converted into chemical neurotransmitters that travel along the axon of the next neurone.
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Worked solution
When an electrical impulse reaches the end of a pre-synaptic neurone, it causes neurotransmitters to be released from vesicles into the synaptic cleft. These neurotransmitter molecules diffuse across the gap and bind to specific receptors on the post-synaptic membrane, which triggers a new electrical impulse in the next neurone.
Marking scheme
1 mark for the correct answer C. Accept C. Reject all other responses.
Question 12 · multiple_choice
1 marks
In a food chain, the biomass of the producer is \( 8500\text{ kg} \) and the biomass of the primary consumer is \( 935\text{ kg} \). What is the percentage efficiency of biomass transfer from the producer to the primary consumer?
A.\( 0.11\% \)
B.\( 9.1\% \)
C.\( 11.0\% \)
D.\( 90.9\% \)
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Worked solution
The percentage efficiency of biomass transfer is calculated as: \( \text{Efficiency} = \frac{\text{Biomass in primary consumer}}{\text{Biomass in producer}} \times 100 \). Substituting the values: \( \text{Efficiency} = \frac{935\text{ kg}}{8500\text{ kg}} \times 100 = 11\% \).
Marking scheme
1 mark for the correct answer C. Accept C. Reject all other responses.
Question 13 · multiple_choice
1 marks
A student investigates the rate of photosynthesis of an aquatic plant at different light intensities and two different carbon dioxide concentrations, 0.04% and 0.4%. The temperature is kept constant at 20°C. At high light intensity, the rate of photosynthesis is significantly higher at 0.4% carbon dioxide than at 0.04% carbon dioxide. What is the limiting factor for photosynthesis at high light intensity when the carbon dioxide concentration is 0.04%?
A.Light intensity
B.Carbon dioxide concentration
C.Temperature
D.Water availability
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Worked solution
At high light intensity, light is no longer the limiting factor. Because increasing the carbon dioxide concentration from 0.04% to 0.4% increases the rate of photosynthesis, the concentration of carbon dioxide was the factor limiting the rate at the lower concentration of 0.04%.
Marking scheme
Award 1 mark for the correct option B. Reject all other options.
Question 14 · multiple_choice
1 marks
Which of the following correctly describes the sequence of events when a nerve impulse reaches a synapse?
A.Neurotransmitters are released from the post-synaptic neurone and diffuse across the gap to bind to receptors on the pre-synaptic neurone.
B.Electrical impulses jump directly across the synaptic gap to trigger a receptor potential in the post-synaptic membrane.
C.Neurotransmitters are released from the pre-synaptic neurone, diffuse across the gap, and bind to receptors on the post-synaptic neurone.
D.Receptors on the pre-synaptic neurone release electrical signals that travel through the synaptic gap to the post-synaptic neurone.
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Worked solution
When an electrical impulse reaches the end of the pre-synaptic neurone, it triggers the release of chemical neurotransmitters. These molecules diffuse across the synaptic cleft (gap) and bind to specific receptor molecules on the membrane of the post-synaptic neurone, initiating a new electrical impulse.
Marking scheme
Award 1 mark for the correct option C. Reject all other options.
Question 15 · multiple_choice
1 marks
A patient has coronary heart disease caused by the build-up of fatty deposits in their coronary arteries. Which treatment would be most suitable to directly widen the narrowed artery and restore blood flow to the heart muscle?
A.Statins to reduce blood cholesterol levels.
B.A stent inserted into the coronary artery.
C.Heart transplant surgery.
D.Treatment with monoclonal antibodies to target the fatty deposits.
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Worked solution
A stent is a metal mesh tube inserted into a blocked or narrowed coronary artery to physically hold it open, directly restoring blood flow. Statins help prevent further build-up by lowering blood cholesterol but do not physically widen the artery instantly. Heart transplants are used for end-stage heart failure, and monoclonal antibodies are not a treatment for coronary fatty deposits.
Marking scheme
Award 1 mark for the correct option B. Reject all other options.
Section B
Answer all questions. Quality of extended response will be assessed in questions marked with an asterisk (*).
16 Question · 73.5 marks
Question 1 · Structured
4 marks
Explain how the body responds when the water potential of the blood decreases (becomes too concentrated) during exercise. In your answer, refer to the pituitary gland, ADH, and the kidney tubules.
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Worked solution
When blood water potential decreases (for example, due to sweating during exercise), the change is detected by osmoreceptors in the hypothalamus of the brain. The hypothalamus signals the pituitary gland to release more Antidiuretic Hormone (ADH) into the blood. ADH travels to the kidneys, where it acts on the collecting ducts (kidney tubules), making their walls more permeable to water. As a result, more water is reabsorbed back into the blood by osmosis. This conserves water in the body, leading to the production of a smaller volume of highly concentrated urine, and restoring the blood water potential back to normal via negative feedback.
Marking scheme
1. Pituitary gland releases more ADH (following detection of low water potential by the hypothalamus) [1 mark] 2. ADH travels in the bloodstream to the kidney tubules / collecting ducts [1 mark] 3. Permeability of the collecting duct walls to water is increased [1 mark] 4. More water is reabsorbed (by osmosis) back into the blood, producing a lower volume of more concentrated urine [1 mark]
Accept reverse arguments for high water potential if clearly contrasted, but the response must address the decrease to score full marks.
Question 2 · Structured
5 marks
A student investigates the effect of light intensity on the rate of photosynthesis in pondweed. They place the light source at a distance of 10 cm from the pondweed and measure the rate of photosynthesis. They then move the light source to a distance of 30 cm.
(a) Calculate the relative light intensity at 30 cm compared to 10 cm. Show your working using the inverse square law.
(b) Explain the effect that moving the light source to 30 cm will have on the rate of photosynthesis, assuming light intensity is the limiting factor.
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Worked solution
(a) According to the inverse square law, light intensity is inversely proportional to the square of the distance: \( I \propto \frac{1}{d^2} \). At 10 cm: \( I \propto \frac{1}{10^2} = \frac{1}{100} = 0.01 \). At 30 cm: \( I \propto \frac{1}{30^2} = \frac{1}{900} \approx 0.00111 \). Alternatively, the distance has increased by a factor of 3 (since \( 30 / 10 = 3 \)). Since light intensity is inversely proportional to the square of the distance, the intensity decreases by \( 3^2 = 9 \) times. Thus, the light intensity is \( \frac{1}{9} \) (or approximately 11.1%) of the original intensity.
(b) Moving the light source further away significantly decreases the light intensity. Since light intensity is the limiting factor, a decrease in light intensity means less light energy is absorbed by chlorophyll in the chloroplasts. Consequently, the rate of the light-dependent stage of photosynthesis decreases, reducing the overall rate of glucose and oxygen production.
Marking scheme
Part (a) [3 marks total]: - Calculates that distance has increased by a factor of 3 (\( 30 / 10 = 3 \)) OR calculates \( d^2 \) values of 100 and 900 [1 mark]. - Uses the inverse square relationship (\( 1/d^2 \) or divides by \( 3^2 \)) [1 mark]. - Correct final answer: decreases by a factor of 9 OR is \( \frac{1}{9} \) (accept 0.11 or 11%) of the original value [1 mark].
Part (b) [2 marks total]: - Identifies that the rate of photosynthesis will decrease [1 mark]. - Explains that less light energy is absorbed by chlorophyll (or less energy is available for chemical reactions/synthesis of glucose) [1 mark].
Question 3 · Structured
5 marks
Many species of bacteria have evolved resistance to antibiotics. Explain how natural selection has led to the development of populations of antibiotic-resistant bacteria, and suggest one way that doctors can help prevent further resistance from developing.
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Worked solution
Antibiotic resistance arises due to natural selection. Within any bacterial population, mutations occasionally occur during DNA replication. This can produce a new allele that makes a bacterium resistant to a specific antibiotic. When the host organism is treated with this antibiotic, it acts as a selective pressure. The non-resistant bacteria are killed, but the resistant bacterium survives (survival of the fittest). Without competition from other bacteria, the surviving resistant bacterium reproduces rapidly via binary fission, passing the allele for resistance on to all its offspring. Over time, and with repeated exposure to the antibiotic, the frequency of the resistant allele increases until the entire population is resistant. To prevent this, doctors should avoid over-prescribing antibiotics (such as for viral infections like colds) and advise patients to always complete their full prescribed course to ensure all bacteria are eradicated.
Marking scheme
Explain how natural selection leads to resistance [4 marks max]: - Mutation occurs in DNA of a bacterium creating a resistant allele / variation [1 mark]. - Antibiotic acts as a selection pressure / kills non-resistant bacteria [1 mark]. - Resistant bacteria survive (survival of the fittest) and reproduce (by binary fission) [1 mark]. - Resistant allele is passed on to offspring / next generation [1 mark]. - Over time, the proportion/frequency of resistant bacteria in the population increases [1 mark].
Prevention method [1 mark max]: - Only prescribe antibiotics when necessary / do not prescribe for viral/mild infections [1 mark] OR advise patients to complete the entire course of antibiotics [1 mark].
Question 4 · Structured Short Answer
4.5 marks
A student investigates the rate of photosynthesis in pondweed.
(a) Explain why measuring the volume of oxygen gas produced using a gas syringe is a more accurate method than counting the number of bubbles released per minute. [2 marks]
(b) The light source is placed at a distance of \( 15\text{ cm} \) from the pondweed. It is then moved to a distance of \( 45\text{ cm} \).
Calculate the relative light intensity at \( 45\text{ cm} \) as a fraction of the intensity at \( 15\text{ cm} \), using the inverse square law: \(\text{Light Intensity} \propto \frac{1}{\text{distance}^2}\). Show your working. [2.5 marks]
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Worked solution
Part (a): - Counting bubbles is less accurate because bubbles can vary in volume/size. - Some bubbles might be released too quickly to count accurately, whereas a gas syringe provides a precise, quantitative measurement of volume.
Part (b): - The distance increases by a factor of \( \frac{45}{15} = 3 \). - According to the inverse square law, light intensity is inversely proportional to the square of the distance. - Therefore, the intensity decreases by a factor of \( 3^2 = 9 \). - The relative light intensity at \( 45\text{ cm} \) compared to \( 15\text{ cm} \) is \( \frac{1}{9} \) (or approximately \( 0.11 \)).
Marking scheme
Part (a) [Max 2 marks]: - 1 mark: For stating that bubbles can be of different sizes/volumes. - 1 mark: For stating that counting is prone to human error (e.g., miscounting very fast bubbles) / gas syringe gives precise quantitative volume measurement.
Part (b) [Max 2.5 marks]: - 1 mark: For determining that the distance increases by a factor of 3 (or calculating individual raw relative intensities: \( \frac{1}{15^2} = \frac{1}{225} \) and \( \frac{1}{45^2} = \frac{1}{2025} \)). - 1 mark: For squaring the scale factor (e.g., \( 3^2 = 9 \) or showing the division \( \frac{225}{2025} \)). - 0.5 marks: For the correct final fraction of \( \frac{1}{9} \) or decimal \( 0.11 \) (accept \( 0.111... \)).
Question 5 · Structured Short Answer
4.5 marks
A student uses a potometer to estimate the rate of transpiration of a leafy shoot.
(a) The capillary tube of the potometer has a cross-sectional area of \( 0.6\text{ mm}^2 \). The air bubble moves a distance of \( 35\text{ mm} \) in a time of \( 10\text{ minutes} \).
Calculate the rate of water uptake by the shoot in \(\text{mm}^3\text{ per minute}\). Show your working. [2.5 marks]
(b) Explain how placing the apparatus in an environment with high humidity would affect the rate of transpiration. [2 marks]
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Worked solution
Part (a): - First, calculate the volume of water taken up: \( \text{Volume} = \text{cross-sectional area} \times \text{distance} = 0.6\text{ mm}^2 \times 35\text{ mm} = 21\text{ mm}^3 \). - Next, calculate the rate of water uptake per minute: \( \text{Rate} = \frac{\text{Volume}}{\text{Time}} = \frac{21\text{ mm}^3}{10\text{ minutes}} = 2.1\text{ mm}^3\text{/min} \).
Part (b): - High humidity increases the concentration of water vapour in the air surrounding the leaf. - This reduces the concentration gradient of water vapour between the inside of the leaf and the external air. - As a result, the rate of diffusion/transpiration decreases.
Marking scheme
Part (a) [Max 2.5 marks]: - 1 mark: Correct calculation of volume of water uptake (\( 0.6 \times 35 = 21\text{ mm}^3 \)). - 1 mark: Dividing the calculated volume by the time (\( 21 / 10 \)). - 0.5 marks: Correct final value of \( 2.1 \).
Part (b) [Max 2 marks]: - 1 mark: Stating that transpiration rate decreases. - 1 mark: Explaining that high humidity reduces the concentration gradient / diffusion gradient of water vapour between the inside of the leaf and the outside air.
Question 6 · Structured Short Answer
4.5 marks
In pea plants, the allele for round seeds (\(R\)) is dominant to the allele for wrinkled seeds (\(r\)). A heterozygous round-seeded plant is crossed with a wrinkled-seeded plant.
(a) State the genotype of the wrinkled-seeded plant and explain your answer. [2 marks]
(b) Complete a genetic cross to determine the probability of obtaining offspring with wrinkled seeds. State this probability as a percentage. [2.5 marks]
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Worked solution
Part (a): - The genotype of the wrinkled-seeded plant is \( rr \). - Wrinkled seeds are a recessive trait, which means the phenotype is only expressed when two copies of the recessive allele are present (homozygous recessive).
Part (b): - The heterozygous round plant has the genotype \( Rr \). - The wrinkled plant has the genotype \( rr \). - Crossing them (\( Rr \times rr \)) yields the following offspring genotypes: \( Rr \) (round) and \( rr \) (wrinkled) in a 1:1 ratio. - The probability of obtaining wrinkled offspring (\( rr \)) is \( \frac{2}{4} = 50\% \).
Marking scheme
Part (a) [Max 2 marks]: - 1 mark: Correct genotype \( rr \). - 1 mark: Explanation that wrinkled is a recessive trait / recessive alleles are only expressed when homozygous / when no dominant allele is present.
Part (b) [Max 2.5 marks]: - 1 mark: Correct identification of parent genotypes/gametes as \( Rr \) and \( rr \). - 1 mark: Correct cross diagram (Punnett square) showing offspring genotypes (\( Rr \) and \( rr \)). - 0.5 marks: Correct probability stated as \( 50\% \).
Question 7 · Structured Short Answer
4 marks
A student investigated the effect of pH on the rate of starch breakdown by amylase. At pH 5, it took 120 seconds for the starch to be completely broken down. At pH 7, it took 40 seconds for the starch to be completely broken down. (a) Calculate the rate of amylase activity at pH 7. Use the formula: \(\text{Rate} = \frac{1}{\text{time}}\). Give your answer to 3 decimal places and state the correct unit. [2] (b) Explain, in terms of active site shape, why the rate of reaction was slower at pH 5 than at pH 7. [2]
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Worked solution
For part (a), the calculation is \(\text{Rate} = \frac{1}{40\text{ s}} = 0.025\). The unit of rate when time is measured in seconds is per second (\(\text{s}^{-1}\)). For part (b), enzyme activity is highly dependent on pH. Since pH 7 is closer to the optimum pH of amylase, the active site maintains its complementary shape to the starch substrate. At pH 5, the change in hydrogen ion concentration alters the intermolecular bonds within the enzyme, changing the shape of the active site so that fewer enzyme-substrate complexes can form.
Marking scheme
Part (a): [2 marks total] - 1 mark for correct calculation of 0.025 (accept 1/40). - 1 mark for correct unit: \(\text{s}^{-1}\) or per second or 1/s (reject 'seconds' or 's').
Part (b): [2 marks total] - 1 mark for stating that the active site changes shape / enzyme is denatured at pH 5. - 1 mark for explaining that the substrate is no longer complementary (or cannot bind / fewer enzyme-substrate complexes form).
Question 8 · Structured Short Answer
5 marks
When a person moves from a warm room into a cold environment, their body initiates several physiological responses to maintain a constant internal core body temperature. (a) Describe how the body detects a decrease in external temperature. [1] (b) Explain how vasoconstriction and shivering help to maintain a core body temperature of approximately \(37\text{ }^\circ\text{C}\) in a cold environment. [4]
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Worked solution
Part (a): Temperature receptors in the skin act as sensory receptors that detect external temperature changes and send nerve impulses to the thermoregulatory centre in the hypothalamus. Part (b): In cold conditions, the muscles in the walls of arterioles supplying skin capillaries contract. This vasoconstriction redirects blood flow away from the skin surface to the core, minimizing heat loss via radiation. Additionally, involuntary rapid contractions of skeletal muscles (shivering) increase the rate of aerobic respiration, which is an exothermic reaction, releasing thermal energy to raise the temperature of the blood.
Marking scheme
Part (a): [1 mark total] - 1 mark for temperature receptors (or thermoreceptors) in the skin.
Part (b): [4 marks total] - 1 mark for identifying that arterioles constrict/narrow (do not accept 'capillaries constrict'). - 1 mark for explaining that vasoconstriction reduces blood flow close to the skin surface, reducing heat loss by radiation/convection. - 1 mark for identifying shivering as rapid muscle contraction and relaxation. - 1 mark for explaining that respiration in muscle cells releases heat/thermal energy.
Question 9 · Structured Short Answer
5 marks
Hospital-acquired infections caused by antibiotic-resistant bacteria, such as MRSA, are a major concern in modern medicine. (a) State how antibiotic resistance arises in a bacterial population before the population is exposed to the antibiotic. [1] (b) Explain how natural selection leads to an increase in the proportion of antibiotic-resistant bacteria in a hospital when patients are treated with antibiotics. [4]
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Worked solution
Part (a): Antibiotic resistance originates from random, spontaneous mutations that occur in the bacterial genome during DNA replication. Part (b): When a population of bacteria is exposed to an antibiotic, the antibiotic acts as an environmental selection pressure. The normal, sensitive bacteria are destroyed, reducing competition. The mutated, resistant bacteria survive the treatment. Since they are still alive, they multiply rapidly by binary fission. They pass the gene/allele responsible for resistance to their offspring, which dramatically increases the frequency of the resistant phenotype within the hospital environment over time.
Marking scheme
Part (a): [1 mark total] - 1 mark for identifying random/spontaneous mutation in DNA/genes.
Part (b): [4 marks total] - 1 mark for identifying that the antibiotic acts as a selection pressure / non-resistant bacteria are killed while resistant bacteria survive. - 1 mark for stating that the surviving resistant bacteria reproduce (accept multiply / divide / binary fission). - 1 mark for stating that they pass on the gene/allele for resistance to their offspring. - 1 mark for stating that this increases the proportion/frequency of the resistant bacteria in the population over time.
Question 10 · Structured Short Answer
4 marks
A student investigates how nitrate ions are absorbed by barley root hair cells.
They measure the rate of nitrate absorption under two conditions: - **Condition A**: Normal root hair cells. - **Condition B**: Root hair cells treated with a metabolic poison that inhibits aerobic respiration.
Explain how and why the absorption of nitrate ions differs between Condition A and Condition B.
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Worked solution
1. In Condition A, nitrate ions are absorbed by active transport. This process moves ions from a lower concentration in the soil to a higher concentration inside the root hair cells (against the concentration gradient). 2. Active transport requires energy in the form of ATP, which is produced during aerobic respiration. 3. In Condition B, the metabolic poison stops aerobic respiration, meaning no ATP is produced. 4. Consequently, active transport cannot take place in Condition B. The rate of nitrate absorption will drop significantly, as the ions can now only enter slowly via passive diffusion down a concentration gradient (if one exists).
Marking scheme
Award 1 mark for each of the following points, up to a maximum of 4 marks: - **MP1**: Identifies that in Condition A, nitrate ions are absorbed via active transport / against their concentration gradient (1). - **MP2**: Explains that active transport requires energy / ATP from aerobic respiration (1). - **MP3**: Explains that in Condition B, the inhibition of respiration means no ATP is produced (1). - **MP4**: Concludes that absorption is significantly reduced / stops because active transport cannot occur without ATP / only slow passive diffusion occurs (1).
Question 11 · Structured Short Answer
4 marks
In tomato plants, resistance to a wilt fungus is controlled by a single gene. The allele for resistance (\(R\)) is dominant to the allele for susceptibility (\(r\)).
A plant breeder crosses a heterozygous resistant tomato plant with a susceptible tomato plant.
1. State the genotypes of both parent plants. 2. Complete a genetic cross explanation (or Punnett square) to determine the expected ratio of resistant to susceptible offspring.
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Worked solution
1. The heterozygous resistant parent must have one dominant and one recessive allele, so its genotype is \(Rr\). The susceptible parent is recessive, so its genotype must be homozygous recessive, \(rr\). 2. The gametes produced by the heterozygous parent are \(R\) and \(r\). The gametes produced by the susceptible parent are all \(r\). 3. Crossing these gives the following offspring genotypes: - \(Rr\) (resistant) - \(rr\) (susceptible) 4. The expected phenotypic ratio is therefore 1 resistant to 1 susceptible (or a 1:1 ratio / 50% chance of each phenotype).
Marking scheme
Award marks as follows: - **MP1**: Correct parent genotypes: heterozygous is \(Rr\) and susceptible is \(rr\) (1). - **MP2**: Correct identification of gametes: \(R\) and \(r\) from one parent, \(r\) (or \(r\) and \(r\)) from the other (1). - **MP3**: Correct offspring genotypes shown (\(Rr\) and \(rr\)) in a Punnett square or description (1). - **MP4**: Correct phenotypic ratio given as 1:1, 50% / 50%, or 1 resistant to 1 susceptible (1).
*Accept: alternative clear symbols only if defined, but penalize 1 mark if R and r are mixed up.*
Question 12 · Structured Short Answer
5 marks
A person eats a meal rich in carbohydrates, causing their blood glucose level to rise.
Explain how the endocrine system coordinates the homeostatic response to return the blood glucose level back to normal. Refer to the organs, hormones, and target tissues involved.
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Worked solution
1. The rise in blood glucose level is detected by cells in the pancreas. 2. The pancreas responds by secreting the hormone insulin into the bloodstream. 3. Insulin travels through the circulatory system to its target organs, which are the liver and muscle cells. 4. Insulin causes these target cells to absorb glucose from the blood and convert it into insoluble glycogen for storage. 5. This removal of glucose from the blood causes the blood glucose level to fall back to its normal set point.
Marking scheme
Award 1 mark for each of the following points, up to a maximum of 5 marks: - **MP1**: States that the pancreas detects the high blood glucose level (1). - **MP2**: States that the pancreas secretes/releases the hormone insulin (1). - **MP3**: Identifies that insulin travels in the blood / to the liver or muscle cells (target organs) (1). - **MP4**: Explains that insulin causes liver/muscle cells to take in glucose / convert glucose to glycogen (1). - **MP5**: States that this results in blood glucose levels decreasing / returning to normal (1).
*Reject: Glucagon or glycogen storage in brain/other organs for MP4.*
Question 13 · structured
4 marks
In a deciduous woodland, the biomass of oak trees is \(12000\text{ kg}\). The biomass of the caterpillars feeding on them is \(1440\text{ kg}\). The biomass of blue tits feeding on caterpillars is \(108\text{ kg}\).
a) Calculate the percentage efficiency of biomass transfer from the oak trees to the caterpillars. Show your working.
b) Explain two reasons why not all the biomass from the oak trees is transferred to the biomass of the caterpillars.
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Worked solution
a) To calculate the percentage efficiency of biomass transfer:
\(\text{Efficiency} = \frac{\text{Biomass transferred to caterpillars}}{\text{Biomass of oak trees}} \times 100\)
b) Only a small proportion of biomass is converted into new caterpillar tissues because: 1. Not all parts of the oak trees are ingested (e.g. roots and woody bark are left behind). 2. Some ingested material cannot be digested and is lost as waste in feces (egestion). 3. Caterpillar cells respire, using up glucose (biomass) to release energy, releasing carbon dioxide and water as waste.
Marking scheme
Part a: [2 marks total] - 1 mark for correct working: \(\frac{1440}{12000} \times 100\) (or equivalent) - 1 mark for correct final answer: \(12\%\) (Accept 12 without % sign if unit is assumed)
Part b: [2 marks total - 1 mark for each valid reason explained up to 2] - Not all parts of the oak are eaten/digested (e.g. woody parts) [1 mark] - Biomass is lost as waste during respiration (as carbon dioxide and water) [1 mark] - Biomass is lost via egestion / in feces / excretion [1 mark]
Question 14 · structured
4 marks
Root hair cells absorb mineral ions from the soil. An investigation measured the rate of potassium ion absorption by plant roots in different oxygen conditions. The results are shown below:
• With oxygen: \(4.8\text{ arbitrary units (a.u.)}\) • Without oxygen: \(0.6\text{ arbitrary units (a.u.)}\)
a) Explain why the rate of potassium ion absorption is much lower when oxygen is absent. Use ideas about respiration and transport processes in your answer.
b) Describe one structural adaptation of root hair cells that increases the rate of absorption of water and mineral ions.
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Worked solution
a) Active transport is used to move ions against their concentration gradient from the soil into the root hair cells. This process requires energy (in the form of ATP) provided by cellular respiration. Aerobic respiration requires oxygen to produce large amounts of ATP. When oxygen is absent, cells must rely on anaerobic respiration, which is much less efficient and yields significantly less energy, slowing down active transport and ion uptake.
b) Root hair cells are adapted to their function by having a long hair-like extension that significantly increases the surface area for diffusion and active transport. They also have thin cell walls to reduce the diffusion distance, and contain many mitochondria to supply the ATP needed for active transport.
Marking scheme
Part a: [3 marks total] - 1 mark for identifying that potassium ions are absorbed by active transport (against a concentration gradient). - 1 mark for stating that active transport requires energy / ATP (from respiration). - 1 mark for explaining that aerobic respiration requires oxygen, so without oxygen only anaerobic respiration occurs which yields much less energy.
Part b: [1 mark total] - 1 mark for describing a valid adaptation: long hair-like extension to increase surface area OR thin cell wall to minimize diffusion distance OR high concentration of mitochondria to synthesize ATP.
Question 15 · structured
5 marks
a) Complete the comparison of Type 1 and Type 2 diabetes by describing the primary biological cause of each condition.
Type 1 cause: Type 2 cause:
b) Explain how the hormone glucagon regulates blood glucose levels when they fall below normal levels, and describe why this mechanism is an example of negative feedback.
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Worked solution
a) Type 1 diabetes is an autoimmune disease where the pancreas stops producing insulin. Type 2 diabetes is a condition where the pancreas still produces insulin, but the body cells have become resistant to it and fail to respond properly.
b) When blood glucose levels fall (e.g., during exercise or fasting), the pancreas detects this change and secretes the hormone glucagon into the blood. Glucagon targets liver and muscle cells, causing them to convert stored glycogen back into glucose, which is then released into the bloodstream. This raises blood glucose levels back to the normal set-point. This is an example of negative feedback because the physiological response (raising glucose levels) acts to reverse and counteract the initial stimulus (the drop in blood glucose).
Marking scheme
Part a: [2 marks total] - 1 mark for stating that Type 1 is caused by the pancreas failing to produce insulin (due to destruction of beta cells). - 1 mark for stating that Type 2 is caused by body cells failing to respond to insulin / becoming resistant to insulin.
Part b: [3 marks total] - 1 mark for explaining that the pancreas detects low blood glucose and secretes glucagon. - 1 mark for explaining that glucagon triggers the breakdown of glycogen into glucose (in the liver/muscles) which enters the blood. - 1 mark for explaining that this is negative feedback because the response reverses/counteracts the initial drop to return glucose to the normal set-point.
Question 16 · Extended Response
6 marks
A runner is training on a hot summer day. As they run, their body temperature begins to rise above normal.
* Describe and explain how the human body detects and coordinates the response to this rise in body temperature, and explain the mechanisms used by the skin to return the body temperature back to normal.
In your answer, you should refer to: - how the change in temperature is detected and processed - the physical responses of the skin to help cool the body.
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Worked solution
### Detection and Coordination: - **Detection:** Temperature receptors in the skin detect changes in the external temperature, while temperature receptors in the brain's hypothalamus detect changes in the temperature of the blood flowing through it. - **Coordination:** The hypothalamus acts as the processing/coordination centre. It receives this sensory information and sends electrical impulses along motor neurones to the effectors (muscles and glands in the skin) to initiate cooling mechanisms.
### Skin Effector Mechanisms: - **Vasodilation:** Arterioles supplying blood to the capillaries near the surface of the skin dilate (widen). This allows a greater volume of warm blood to flow closer to the surface of the skin, increasing the rate of heat loss to the surrounding air by radiation. (Note: capillaries themselves do not widen or move closer to the skin). - **Sweating:** Sweat glands in the dermis are stimulated to secrete sweat onto the surface of the skin. As the water in sweat evaporates, it absorbs and takes away thermal energy (latent heat of vaporisation) from the body, cooling it down. - **Flattening of Hairs (Pilorelaxation):** Hair erector muscles in the skin relax, causing the skin hairs to lie flat. This prevents a layer of still, warm air from being trapped close to the skin surface, allowing better convection and heat loss.
Marking scheme
### Level 3 (5–6 marks) - **Criteria:** Detailed and accurate description of both detection/coordination and at least two skin cooling mechanisms (vasodilation and sweating) explained in detail with correct terminology. - **Key requirements:** Identifies the hypothalamus as the coordination centre detecting blood temperature. Clearly explains that during vasodilation, arterioles dilate (not capillaries) to increase blood flow near the surface. Explains that sweat cools the body via evaporation. - **Quality of Communication:** There is a well-developed, line of reasoning which is clear and logically structured. The information presented is relevant and substantiated.
### Level 2 (3–4 marks) - **Criteria:** Describes both detection/coordination and at least one skin mechanism, or describes multiple skin mechanisms with less focus on detection/coordination. - **Key requirements:** Mentions the brain or hypothalamus detecting temperature. Describes sweating or vasodilation with some scientific accuracy (e.g., understands sweat evaporates or that blood flow to the skin increases). - **Quality of Communication:** There is a line of reasoning with some structure. The information presented is in the most part relevant with some appropriate use of specialist terms.
### Level 1 (1–2 marks) - **Criteria:** Identifies basic points about detection or names skin mechanisms (e.g., sweating, vasodilation, hairs lying flat) without detailed explanations of how they work. - **Key requirements:** Simple statements such as 'the brain detects temperature rise' or 'you sweat to cool down'. - **Quality of Communication:** Information is basic and lacks structure. Answers may contain errors or lack specialist terminology.
### 0 marks - No response or no response worthy of credit.
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