An original Thinka practice paper modelled on the structure and difficulty of the Jun 2022 AQA AS Level Biology 7401 paper. Not affiliated with or reproduced from AQA.
卷一
Answer all questions. Show all your working. Write your answers in the spaces provided.
28 題目 · 74.19999999999999 分
題目 1 · Recall
2.2 分
An electron micrograph shows a chloroplast with a length of \( 48\text{ mm} \). The magnification of the image is \( \times 8000 \). Calculate the actual length of the chloroplast in micrometres (\(\mu\text{m}\)). Show your working.
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解題
First, convert the image length from millimetres to micrometres: \( 48\text{ mm} \times 1000 = 48000\ \mu\text{m} \). Next, use the magnification formula: \( \text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}} \). Therefore, \( \text{Actual Size} = \frac{48000}{8000} = 6\ \mu\text{m} \).
評分準則
1 mark for converting image size correctly to \( 48000\ \mu\text{m} \) OR for correctly rearranging the equation as \( \text{Actual} = \frac{\text{Image}}{\text{Magnification}} \). 1.2 marks for the correct final answer of 6.
題目 2 · Recall
2.2 分
Describe the process of cooperative binding of oxygen to haemoglobin and explain how this affects the shape of the oxygen dissociation curve.
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解題
The binding of the first oxygen molecule to haemoglobin induces a conformational (shape) change in its quaternary structure. This change exposes the remaining binding sites on the haem groups, making it significantly easier for the second and third oxygen molecules to bind. This cooperative binding results in a steep rise in oxygen saturation over a narrow range of partial pressures, producing the characteristic sigmoid (S-shaped) oxygen dissociation curve.
評分準則
1 mark for stating that the binding of the first oxygen molecule alters the shape/conformation of the haemoglobin molecule. 1.2 marks for explaining that this shape change exposes/uncovers other binding sites, facilitating the binding of subsequent oxygen molecules (producing a sigmoid curve).
題目 3 · Recall
2.2 分
Explain how companion cells are structurally and biochemically adapted to facilitate the active loading of sucrose into sieve tube elements during translocation.
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解題
Companion cells contain a very high density of mitochondria which produce ATP via aerobic respiration. This ATP is used to actively transport hydrogen ions out of the companion cells into the surrounding cell wall. This establishes a high concentration gradient of hydrogen ions, which then diffuse back into the companion cell through co-transporter proteins, bringing sucrose molecules with them against their concentration gradient into the sieve tube elements.
評分準則
1 mark for identifying the high concentration of mitochondria that produce ATP for the active transport of hydrogen ions. 1.2 marks for explaining the role of co-transporter proteins in allowing hydrogen ions and sucrose to enter the cells together.
題目 4 · Recall
2.2 分
Explain how the counter-current exchange system in fish gills ensures maximum efficiency of oxygen absorption from water into the blood.
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解題
In the counter-current system, blood flows through the capillaries in the opposite direction to the flow of water over the gill lamellae. This arrangement ensures that water with a relatively high oxygen concentration always meets blood with a lower oxygen concentration. Consequently, a concentration/diffusion gradient for oxygen is maintained along the entire length of the gill lamella, maximizing the amount of oxygen that diffuses into the blood.
評分準則
1 mark for stating that blood and water flow in opposite directions across the gill lamellae. 1.2 marks for explaining that this maintains a concentration gradient of oxygen across the entire length of the capillary/lamella.
題目 5 · Recall
2.2 分
During periods of high activity, such as flight, insects produce lactate through anaerobic respiration in their muscles. Explain how the production of lactate helps to increase the rate of oxygen diffusion into their tissues.
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解題
Lactate is a soluble solute that accumulates in the muscle tissue during anaerobic respiration. This lowers the water potential of the muscle cells. Consequently, water moves from the ends of the tracheoles into the muscle cells by osmosis down a water potential gradient. This withdrawal of fluid lowers the fluid level in the tracheoles, drawing air (and oxygen) deeper and closer to the active muscle cells, shortening the diffusion distance.
評分準則
1 mark for explaining that lactate lowers the water potential of the muscle cells, causing water to move from the tracheoles into the cells by osmosis. 1.2 marks for explaining that this decreases fluid level in the tracheoles, drawing air closer to the cells / shortening the diffusion distance.
題目 6 · Recall
2.2 分
Describe how a macrophage acts as an antigen-presenting cell (APC) to activate helper T cells (\(\text{T}_h\) cells).
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解題
A macrophage engulfs a pathogen via phagocytosis and uses lysosomal hydrolytic enzymes to break it down. It then processes the pathogen's antigens and presents (displays) them on its own cell surface membrane. A specific helper T cell with a complementary receptor binds to these presented antigens, which activates the helper T cell to divide by mitosis.
評分準則
1 mark for engulfing/digesting the pathogen and presenting/displaying the foreign antigens on its own cell surface membrane. 1.2 marks for explaining that helper T cells with complementary receptors bind to these presented antigens to become activated.
題目 7 · Recall
2.2 分
A student prepared a root tip squash to observe mitosis. Out of a total of \( 350 \) cells observed in the field of view, \( 28 \) cells were in prophase, \( 14 \) cells were in metaphase, \( 7 \) cells were in anaphase, and \( 21 \) cells were in telophase. Calculate the mitotic index of this tissue. Show your working.
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解題
First, calculate the total number of cells undergoing mitosis: \( 28 + 14 + 7 + 21 = 70 \) cells. Next, use the mitotic index formula: \( \text{Mitotic Index} = \frac{\text{Number of cells in mitosis}}{\text{Total number of cells}} \). Therefore, \( \text{Mitotic Index} = \frac{70}{350} = 0.2 \) (or \( 20\% \)).
評分準則
1 mark for correctly calculating the number of dividing cells as 70 OR for stating the correct formula. 1.2 marks for the correct final answer of 0.2 (accept 20%).
題目 8 · Recall
2.2 分
Define what is meant by a phylogenetic classification system and describe how taxa are arranged in a classification hierarchy.
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解題
A phylogenetic classification system groups organisms based on their evolutionary history, common ancestry, and shared relationships. A taxonomic hierarchy is a system where organisms are organized into nested groups (taxa) where smaller, more specific groups are placed within larger, more inclusive groups, and there is no overlap between different groups at the same level.
評分準則
1 mark for defining phylogenetic classification as grouping species based on evolutionary relationships / common ancestry. 1.2 marks for describing a hierarchy as containing smaller groups within larger groups, with no overlap.
題目 9 · short-structured
2.2 分
Describe how high hydrostatic pressure at the arteriole end of a capillary leads to the formation of tissue fluid.
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解題
The high hydrostatic pressure at the arteriole end of the capillary is greater than the water potential gradient opposing it. This ultrafiltration forces water, glucose, amino acids, and ions out through the tiny pores in the capillary wall. Large proteins and blood cells are too large to pass through these pores and are retained in the capillary.
評分準則
1. High hydrostatic pressure forces water and small solutes out of the capillary (via ultrafiltration). 2. Large proteins / plasma proteins remain inside the capillary because they are too large to pass through the pores in the capillary wall.
題目 10 · short-structured
2.2 分
Explain how the Bohr effect causes more oxygen to be released to respiring tissues.
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解題
During respiration, tissues release carbon dioxide, which dissolves to form carbonic acid, lowering the pH. This acidic environment alters the tertiary structure of hemoglobin, which reduces its affinity for oxygen. Consequently, oxygen is unloaded more easily at the respiring tissues.
評分準則
1. Increased carbon dioxide concentration lowers blood pH / increases acidity. 2. This alters the tertiary structure / shape of hemoglobin, reducing its affinity for oxygen so it dissociates/unloads more easily.
題目 11 · short-structured
2.2 分
Explain how a concentration gradient of oxygen is maintained between the atmosphere and the tracheoles of an insect during flight.
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解題
Active tissues like muscles consume oxygen rapidly during flight, which lowers the oxygen concentration at the ends of the tracheoles. Concurrently, abdominal pumping (ventilation) physically moves air in and out, maintaining a high concentration of oxygen in the tracheae to preserve a steep diffusion gradient.
評分準則
1. Aerobic respiration in muscle cells consumes oxygen, creating a low oxygen concentration / partial pressure at the tracheoles. 2. Abdominal pumping / muscle contraction ventilates the trachea, bringing in oxygen-rich air from the atmosphere.
題目 12 · short-structured
2.2 分
After a phagocytic cell engulfs a pathogen, a phagosome is formed. Describe the role of lysosomes in the destruction of this pathogen.
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解題
Once the pathogen is enclosed in a phagosome, lysosomes move towards it and fuse with its membrane. This forms a phagolysosome and allows hydrolytic enzymes (lysozymes) to enter the vesicle, where they chemically break down (hydrolyse) the biological molecules of the pathogen.
評分準則
1. Lysosomes fuse with the phagosome membrane (forming a phagolysosome). 2. Lysosomes release lysozymes / hydrolytic enzymes which hydrolyse / digest the pathogen.
題目 13 · short-structured
2.2 分
Explain why a Transmission Electron Microscope (TEM) has a higher resolution than a light microscope.
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解題
The resolution of a microscope is determined by the wavelength of the radiation used. Light microscopes use visible light which has a relatively long wavelength. Transmission Electron Microscopes use a beam of electrons, which has a much shorter wavelength, allowing them to resolve structures that are much closer together.
評分準則
1. TEM uses an electron beam which has a much shorter wavelength than light. 2. A shorter wavelength allows the microscope to resolve / distinguish between two close objects as separate.
題目 14 · short-structured
2.2 分
A student counted 120 cells in a root tip meristem sample. Of these cells, 18 were in prophase, 6 in metaphase, 4 in anaphase, and 2 in telophase. Calculate the mitotic index of this sample as a percentage.
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解題
To find the mitotic index, calculate the total number of cells undergoing mitosis: \(18 + 6 + 4 + 2 = 30\) cells. Divide this by the total number of cells observed: \(30 / 120 = 0.25\). Express this as a percentage by multiplying by 100: \(0.25 \times 100 = 25\%\).
評分準則
1. Correctly calculates total dividing cells as 30, or sets up the ratio \(30 / 120\). 2. Correct final answer of 25% (or 25).
題目 15 · short-structured
2.2 分
Explain what is meant by a phylogenetic classification system.
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解題
Phylogeny refers to the evolutionary relationships between organisms. A phylogenetic classification system groups organisms based on these relationships and their evolutionary history, showing how closely related different species are. It uses a hierarchy of groups (taxa) where smaller groups are contained within larger ones, with no overlap.
評分準則
1. Groups organisms based on evolutionary relationships / common ancestry. 2. Organised in a hierarchy with smaller groups within larger groups, and no overlap between groups.
題目 16 · short-structured
2.2 分
Explain how the counter-current system in the gills of a fish ensures efficient gas exchange.
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解題
In the gills of a bony fish, blood in the capillaries flows in the opposite direction to the flow of water over the lamellae. This counter-current flow ensures that water with a higher oxygen concentration always meets blood with a lower oxygen concentration, maintaining a steep concentration gradient for diffusion across the entire length of the gas exchange surface.
評分準則
1. Blood and water flow in opposite directions across the lamellae / gas exchange surface. 2. This maintains a concentration gradient / diffusion gradient (of oxygen) across the entire length / width of the capillary (preventing equilibrium).
題目 17 · Recall & Short Structured
2 分
Explain how a rightward shift in the oxygen-haemoglobin dissociation curve (the Bohr effect) is advantageous to actively respiring tissues.
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解題
During active respiration, tissues release more carbon dioxide, which dissolves to form carbonic acid, lowering the local pH. This acidic environment alters the tertiary structure of haemoglobin, reducing its affinity for oxygen. Consequently, oxygen is released or unloaded more easily at the respiring tissues where it is urgently required for aerobic respiration.
評分準則
1. Carbon dioxide release reduces pH, lowering haemoglobin's affinity for oxygen (1 mark). 2. Oxygen is unloaded or dissociated more readily to the actively respiring cells/tissues (1 mark).
題目 18 · Recall & Short Structured
2 分
Explain how the counter-current exchange system in fish gills ensures efficient gas exchange.
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解題
In the counter-current exchange system, blood in the gill lamellae flows in the opposite direction to the flow of water. This ensures that water with a higher oxygen concentration is always adjacent to blood with a lower oxygen concentration, maintaining a diffusion gradient for oxygen along the entire length of the capillary bed.
評分準則
1. Water and blood flow in opposite directions across the gill lamellae (1 mark). 2. This maintains a concentration gradient of oxygen across the entire length of the gill/lamellae (preventing equilibrium) (1 mark).
題目 19 · Recall & Short Structured
2 分
Describe the role of lysosomes during the process of phagocytosis.
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解題
Once a pathogen is engulfed by a phagocyte, it is enclosed in a vesicle called a phagosome. Lysosomes move towards and fuse with this phagosome, forming a phagolysosome. The lysosomes release hydrolytic enzymes (lysozymes) into the vesicle, which chemically digest and break down the pathogen.
評分準則
1. Lysosomes fuse with the phagosome / phagocytic vesicle (1 mark). 2. They release lysozymes / hydrolytic enzymes to digest, hydrolyse, or destroy the pathogen (1 mark). Do not accept simple 'break down' without mention of enzymes or hydrolysis.
題目 20 · Recall & Short Structured
2 分
During cell fractionation, plant tissue is homogenised in an extraction medium. Explain why this medium must be kept ice-cold and be isotonic to the plant cells.
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解題
The extraction medium is kept ice-cold to reduce the kinetic energy of enzymes, preventing them from breaking down or damaging the organelles after cell lysis. It is kept isotonic (having the same water potential as the cell contents) to ensure there is no net movement of water by osmosis, preventing the organelles from bursting or shrinking.
評分準則
1. Ice-cold: reduces/stops enzyme activity to prevent self-digestion/damage to organelles (1 mark). 2. Isotonic: maintains equal water potential to prevent osmotic water movement and stop organelles bursting or shrinking (1 mark).
題目 21 · Recall & Short Structured
2 分
Describe the behaviour of chromosomes during the metaphase stage of mitosis.
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解題
During metaphase, chromosomes (each consisting of two sister chromatids joined at a centromere) line up individually along the equator (metaphase plate) of the cell. Spindle fibres, which extend from the poles of the cell, attach to the centromere of each chromosome.
評分準則
1. Chromosomes align or line up along the equator / middle / metaphase plate of the cell (1 mark). 2. Spindle fibres attach to the centromeres (of the chromosomes) (1 mark).
題目 22 · Recall & Short Structured
2 分
Explain how courtship behaviour helps to ensure successful reproduction within a species.
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解題
Courtship behaviour is species-specific, which allows individuals to recognise other members of their own species to prevent interbreeding and ensure fertile offspring are produced. It also allows individuals to identify mates that are sexually mature and receptive, ensuring synchronization of gamete release.
評分準則
1. Allows recognition of members of the same species to prevent interbreeding / ensure fertile offspring (1 mark). 2. Identifies a mate who is receptive / fertile / synchronises mating (1 mark).
題目 23 · Application & Numerical Analysis
4.5 分
At rest, a healthy student has a cardiac output of \(5.0\text{ dm}^3\text{ min}^{-1}\) and a heart rate of \(72\text{ beats min}^{-1}\). During vigorous exercise, the student's heart rate increases by \(95\%\) and their stroke volume increases by \(40\%\). Calculate the student's cardiac output during exercise in \(dm^3\text{ min}^{-1}\). Show your working.
1. Correct calculation of resting stroke volume (\(0.0694\text{ dm}^3\) or \(69.4\text{ cm}^3\)) OR correct multiplier for exercise heart rate (\(1.95\)) and exercise stroke volume (\(1.40\)) (1.5 marks). 2. Correct calculation of exercise heart rate (\(140.4\text{ beats min}^{-1}\)) OR exercise stroke volume (\(0.0972\text{ dm}^3\) or \(97.2\text{ cm}^3\)) (1.5 marks). 3. Correct final answer of \(13.65\text{ dm}^3\text{ min}^{-1}\) (1.5 marks). Accept \(13.7\) if correct rounding from intermediate steps is demonstrated. Reject all other values.
題目 24 · Application & Numerical Analysis
4.5 分
A student prepared a root tip squash of Allium cepa to determine the mitotic index. They recorded the cell state in four different fields of view: Field 1: 42 in interphase, 6 in prophase, 2 in metaphase, 1 in anaphase, 1 in telophase. Field 2: 38 in interphase, 5 in prophase, 3 in metaphase, 2 in anaphase, 0 in telophase. Field 3: 45 in interphase, 4 in prophase, 1 in metaphase, 1 in anaphase, 2 in telophase. Field 4: 41 in interphase, 7 in prophase, 2 in metaphase, 1 in anaphase, 1 in telophase. Calculate the overall mitotic index for the combined data of all four fields of view as a percentage. Give your answer to 3 significant figures. Show your working.
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解題
1. Sum the total number of cells observed across all 4 fields: Total cells = \((42+6+2+1+1) + (38+5+3+2+0) + (45+4+1+1+2) + (41+7+2+1+1) = 52 + 48 + 53 + 52 = 205\) cells. 2. Sum the number of cells undergoing mitosis: Dividing cells = \((6+2+1+1) + (5+3+2+0) + (4+1+1+2) + (7+2+1+1) = 10 + 10 + 8 + 11 = 39\) cells. 3. Calculate the mitotic index: \((39 / 205) \times 100 = 19.024\%\). 4. Round to 3 significant figures: \(19.0\%\).
評分準則
1. Correct calculation of total number of cells observed across all fields (\(205\)) (1.5 marks). 2. Correct calculation of total number of dividing cells (\(39\)) (1.5 marks). 3. Correct calculation of mitotic index to 3 significant figures: \(19.0\%\) (1.5 marks). Accept \(19\%\) or \(19.02\%\) for 1 mark if 3 significant figures rule is ignored.
題目 25 · Application & Numerical Analysis
4.5 分
An electron micrograph shows a transmission electron microscope (TEM) image of a mitochondrion. The magnification of this micrograph is \(\times 30,000\). In this micrograph, the mitochondrion has a measured length of \(7.2\text{ cm}\). A student wants to produce a drawing of this same mitochondrion, but at a higher magnification of \(\times 45,000\). Calculate the length of the mitochondrion in the student's drawing. Give your answer in centimeters (cm). Show your working.
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解題
1. Calculate the actual length of the mitochondrion: Actual length = Image size / Magnification = \(7.2\text{ cm} / 30,000 = 72,000\ \mu\text{m} / 30,000 = 2.4\ \mu\text{m}\). 2. Calculate the length of the student's drawing: Drawing size = Actual length \(\times\) New magnification = \(2.4\ \mu\text{m} \times 45,000 = 108,000\ \mu\text{m}\). 3. Convert back to centimeters: \(108,000\ \mu\text{m} = 108\text{ mm} = 10.8\text{ cm}\). Alternative Method: Ratio of magnification = \(45,000 / 30,000 = 1.5\). Drawing length = \(7.2\text{ cm} \times 1.5 = 10.8\text{ cm}\).
評分準則
1. Correct calculation of the actual length of the mitochondrion (\(2.4\ \mu\text{m}\) or \(0.00024\text{ cm}\)) OR correct ratio of magnification (\(1.5\)) (1.5 marks). 2. Correct calculation of the drawing size in micrometers (\(108,000\ \mu\text{m}\)) or millimeters (\(108\text{ mm}\)) (1.5 marks). 3. Correct final answer of \(10.8\text{ cm}\) (1.5 marks). Reject other values.
題目 26 · Application & Numerical Analysis
4.5 分
A person at rest has a tidal volume of \(0.55\text{ dm}^3\) and a ventilation rate of \(14\text{ breaths min}^{-1}\). During moderate exercise, their pulmonary ventilation rate increases by \(150\%\). Their breathing rate also increases to \(22\text{ breaths min}^{-1}\). Calculate the tidal volume of the person during moderate exercise. Give your answer in \(\text{dm}^3\) to 2 decimal places. Show your working.
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解題
1. Calculate resting pulmonary ventilation rate (PVR): PVR = Tidal volume \(\times\) Breathing rate = \(0.55\text{ dm}^3 \times 14 = 7.7\text{ dm}^3\text{ min}^{-1}\). 2. Calculate exercise PVR: An increase of \(150\%\) means the new value is \(250\%\) of the resting value. Exercise PVR = \(7.7 \times 2.5 = 19.25\text{ dm}^3\text{ min}^{-1}\). 3. Calculate exercise tidal volume: Tidal volume = Exercise PVR / Exercise breathing rate = \(19.25 / 22 = 0.875\text{ dm}^3\). 4. Round to 2 decimal places: \(0.88\text{ dm}^3\).
評分準則
1. Correct calculation of resting PVR (\(7.7\text{ dm}^3\text{ min}^{-1}\)) (1.5 marks). 2. Correct calculation of exercise PVR (\(19.25\text{ dm}^3\text{ min}^{-1}\)) OR correct multiplier of \(2.5\) (1.5 marks). 3. Correct final tidal volume of \(0.88\text{ dm}^3\) (1.5 marks). Note: Allow up to 1.5 marks maximum for a final answer of \(0.53\text{ dm}^3\) (if student incorrectly used a multiplier of \(1.5\) instead of \(2.5\)).
題目 27 · Application & Numerical Analysis
4.5 分
An ecologist sampled a community of trees in an ancient woodland and recorded the abundance of each species: Oak: 12, Birch: 5, Holly: 8, Rowan: 3, Hazel: 2. Calculate the index of diversity (d) for this community using the formula: \(d = \frac{N(N-1)}{\sum n(n-1)}\). Give your answer to 3 significant figures. Show your working.
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解題
1. Calculate total number of organisms (N): \(N = 12 + 5 + 8 + 3 + 2 = 30\). Therefore, \(N(N-1) = 30 \times 29 = 870\). 2. Calculate \(n(n-1)\) for each species: Oak: \(12 \times 11 = 132\); Birch: \(5 \times 4 = 20\); Holly: \(8 \times 7 = 56\); Rowan: \(3 \times 2 = 6\); Hazel: \(2 \times 1 = 2\). 3. Calculate sum of \(n(n-1)\): \(132 + 20 + 56 + 6 + 2 = 216\). 4. Calculate index of diversity (d): \(d = 870 / 216 \approx 4.0277...\). 5. Round to 3 significant figures: \(4.03\).
評分準則
1. Correct calculation of \(N(N-1) = 870\) (1.5 marks). 2. Correct calculation of \(\sum n(n-1) = 216\) (1.5 marks). 3. Correct index of diversity rounded to 3 significant figures: \(4.03\) (1.5 marks). Accept \(4.028\) or \(4.0\) for 1 mark if significant figures rules are violated.
題目 28 · Application & Numerical Analysis
4.5 分
An immunologist is performing an ELISA to measure the concentration of antibodies in a patient's serum. They perform a dilution series: 1. A 1 in 5 dilution of the stock serum to produce Dilution 1. 2. A 1 in 10 dilution of Dilution 1 to produce Dilution 2. 3. A 1 in 4 dilution of Dilution 2 to produce Dilution 3. The concentration of antibody in Dilution 3 is measured to be \(1.25\ \mu\text{g cm}^{-3}\). Calculate the concentration of antibody in the original stock serum. Give your answer in \(\text{mg dm}^{-3}\). Show your working.
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解題
1. Calculate overall dilution factor: \(5 \times 10 \times 4 = 200\). 2. Calculate concentration in original stock serum: \(1.25\ \mu\text{g cm}^{-3} \times 200 = 250\ \mu\text{g cm}^{-3}\). 3. Convert the unit to \(\text{mg dm}^{-3}\): \(1\ \mu\text{g} = 10^{-3}\text{ mg}\) and \(1\text{ cm}^3 = 10^{-3}\text{ dm}^3\). Thus, \(250\ \mu\text{g cm}^{-3} = 250\text{ mg dm}^{-3}\).
評分準則
1. Correct calculation of cumulative dilution factor (\(200\)) (1.5 marks). 2. Correct calculation of concentration in micrograms per cubic centimeter (\(250\ \mu\text{g cm}^{-3}\)) (1.5 marks). 3. Correct final concentration in milligrams per cubic decimeter (\(250\text{ mg dm}^{-3}\)) (1.5 marks). Accept \(250\). Award maximum of 3 marks if dilution factors are added instead of multiplied (dilution factor of \(19\), giving \(23.75\text{ mg dm}^{-3}\)).
卷二
Answer all questions. Show all your working. A scientific calculator and mm ruler are required.
28 題目 · 75.8 分
題目 1 · Recall & Short Structured
2.3 分
Describe and explain the effect of an increase in carbon dioxide concentration on the oxygen dissociation curve of hemoglobin (the Bohr effect).
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解題
At high partial pressures of carbon dioxide (e.g., in rapidly respiring tissues), carbon dioxide dissolves to produce carbonic acid, which dissociates and lowers the pH. The increased acidity alters the tertiary structure of hemoglobin, reducing its affinity for oxygen. Consequently, the oxygen dissociation curve shifts to the right, meaning that hemoglobin releases (unloads) oxygen more readily to meet the high metabolic demands of the tissues.
評分準則
- 1 mark: State that the oxygen dissociation curve shifts to the right (and downwards). - 1 mark: Explain that carbon dioxide lowers the pH, changing the tertiary structure of hemoglobin and reducing its affinity for oxygen. - 0.3 marks: Explain that this results in oxygen being unloaded or released more readily to rapidly respiring tissues.
題目 2 · Recall & Short Structured
2.3 分
Explain how the countercurrent exchange system in the gills of a bony fish ensures efficient gas exchange.
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解題
In bony fish, blood in the gill capillaries flows in the opposite direction to the water flowing over the gill lamellae (countercurrent flow). This arrangement ensures that water with a relatively high oxygen concentration always meets blood with a lower oxygen concentration. Consequently, a diffusion gradient for oxygen is maintained along the entire length of the gill plate, and oxygen continuously diffuses into the blood, preventing the system from reaching equilibrium.
評分準則
- 1 mark: State that water and blood flow in opposite directions (countercurrent flow) across the gill lamellae. - 1 mark: Explain that this maintains an oxygen concentration gradient between the water and the blood across the entire length of the capillary. - 0.3 marks: Clarify that this allows continuous diffusion of oxygen into the blood (equilibrium is never reached).
題目 3 · Recall & Short Structured
2.3 分
Describe two differences between active immunity and passive immunity.
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解題
Active immunity requires exposure to the pathogen or antigen, leading to the production of antibodies by the individual's own plasma cells, as well as the production of memory cells, which provide long-term protection. Passive immunity, on the other hand, involves the direct introduction of antibodies from an external source (e.g., across the placenta or via injection); it does not involve antigen exposure or memory cell production, and the protection is short-term because the external antibodies are eventually broken down.
評分準則
- 1 mark: Active immunity involves exposure to antigens and antibody production by the host, whereas passive immunity involves external antibodies being introduced directly. - 1 mark: Active immunity leads to the production of memory cells and provides long-term immunity, whereas passive immunity does not produce memory cells and only provides short-term protection. - 0.3 marks: For clear contrast presented in a structured, comparative format.
題目 4 · Recall & Short Structured
2.3 分
Contrast the advantages and limitations of a Transmission Electron Microscope (TEM) with a Scanning Electron Microscope (SEM) when imaging cellular structures.
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解題
Transmission Electron Microscopes (TEM) use transmitted electrons to form an image, offering extremely high resolution (resolving power) that allows visualization of the internal ultrastructure of organelles inside cells. However, they require extremely thin specimens and produce only 2D images. Scanning Electron Microscopes (SEM) bounce electrons off the specimen's surface to create 3D images of surface topography, allowing thicker specimens to be used, but they have a lower maximum resolving power than a TEM.
評分準則
- 1 mark: Identify that TEM has a higher resolution/resolving power than SEM and is used to see internal structures, while SEM is used to see 3D surface structures. - 1 mark: Identify that TEM requires extremely thin sections/specimens and produces 2D images, whereas SEM can use thicker specimens. - 0.3 marks: Ensure comparative language is used correctly to contrast both advantages and limitations.
題目 5 · Recall & Short Structured
2.3 分
A student prepared a garlic root tip squash to observe mitosis. In a sample of 240 cells, 18 cells were in prophase, 8 in metaphase, 6 in anaphase, and 4 in telophase. Calculate the mitotic index of this tissue. Show your working.
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解題
To calculate the mitotic index, we divide the number of cells undergoing mitosis by the total number of cells observed. 1. Calculate total cells in mitosis: \(18 \text{ (prophase)} + 8 \text{ (metaphase)} + 6 \text{ (anaphase)} + 4 \text{ (telophase)} = 36\) cells. 2. Total number of cells examined = 240. 3. Mitotic Index = \(\frac{36}{240} = 0.15\) (or \(15\%\)).
評分準則
- 1 mark: Correctly sum the cells in the active stages of mitosis (\(18 + 8 + 6 + 4 = 36\)). - 1 mark: Correctly divide the mitotic cells by the total number of cells (\(\frac{36}{240}\)). - 0.3 marks: Correct final answer of 0.15 or 15% (do not accept 15 without the percentage symbol).
題目 6 · Recall & Short Structured
2.3 分
Explain what is meant by a phylogenetic classification system and define the term 'hierarchy' in this context.
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解題
A phylogenetic classification system arranges species into groups based on their evolutionary origins, shared ancestry, and how closely related they are. A taxonomic hierarchy is a system of classification in which smaller groups (taxa) are nested/contained within larger composite groups (e.g., species within genus, genus within family), and there is no overlap between the groups at any given level.
評分準則
- 1 mark: Define phylogenetic as grouping organisms based on evolutionary relationships, history, or shared ancestors. - 1 mark: Define hierarchy as a system of smaller groups contained within larger groups. - 0.3 marks: State that there is no overlap between the groups at each level of the hierarchy.
題目 7 · Recall & Short Structured
2.3 分
Describe how water moves up the xylem of a plant according to the cohesion-tension theory.
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解題
Water vapor evaporates from the stomata of leaves during transpiration, creating a tension (negative pressure) that pulls water from the xylem vessels into the mesophyll cells. Because water molecules are polar, they form hydrogen bonds with one another; this property is called cohesion. Cohesion allows water molecules to cling together, forming a continuous, unbroken column of water that is pulled up through the xylem under tension. Additionally, adhesion between water molecules and the hydrophilic walls of xylem vessels helps support the water column.
評分準則
- 1 mark: Explain that transpiration/evaporation of water from the leaves creates tension (pulls water up the xylem). - 1 mark: Explain that cohesion (due to hydrogen bonding) between polar water molecules forms a continuous, unbroken column of water. - 0.3 marks: Mention adhesion of water molecules to the walls of the xylem vessels to support the column.
題目 8 · Recall & Short Structured
2.3 分
Explain how the structure of the tracheal system of an insect is adapted to allow efficient gas exchange while minimizing water loss.
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解題
Insects have an internal tracheal system consisting of tracheae and highly branched tracheoles that penetrate directly into respiring tissues. This provides a very large surface area and short diffusion pathway. To prevent excessive water loss, the openings to the exterior (spiracles) can be closed by valves when gas exchange requirements are low. Furthermore, chitin rings support the tracheae to prevent them collapsing under negative pressure, maintaining an open pathway for diffusion.
評分準則
- 1 mark: Explain that highly branched tracheoles provide a large surface area and thin walls provide a short diffusion pathway directly to cells. - 1 mark: State that spiracles have valves/sphincters that can be closed to restrict water loss. - 0.3 marks: Mention that chitin rings keep the trachea open/prevent collapse, or that fluid in the tracheoles can be drawn into tissues during activity to increase the rate of diffusion.
題目 9 · short-answer
2.3 分
A mammal has a heart rate of 75 beats per minute and a stroke volume of 64 cm3. Calculate the cardiac output of this mammal in dm3 min-1. Show your working.
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解題
Cardiac output is calculated using the formula: Cardiac output = heart rate x stroke volume. Heart rate = 75 beats min-1, stroke volume = 64 cm3. Cardiac output = 75 x 64 = 4800 cm3 min-1. To convert cm3 to dm3, divide by 1000: 4800 / 1000 = 4.8 dm3 min-1.
評分準則
1 mark for correct calculation of cardiac output in cm3 min-1 (4800) or correct division by 1000. 1.3 marks for correct final answer of 4.8 dm3 min-1 with correct units.
題目 10 · short-answer
2.3 分
Explain how the counter-current exchange system in the gills of a fish ensures efficient gas exchange.
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解題
In counter-current flow, water and blood flow in opposite directions. This ensures that blood with a lower oxygen concentration always meets water with a higher oxygen concentration, maintaining a concentration gradient for oxygen across the entire length of the capillary, allowing maximum oxygen diffusion into the blood.
評分準則
1 mark for stating that water and blood flow in opposite directions. 1.3 marks for explaining that this maintains a concentration gradient for oxygen across the entire width/length of the lamellae, ensuring diffusion occurs along the whole length.
題目 11 · short-answer
2.3 分
Describe two differences between active immunity and passive immunity.
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解題
Active immunity involves the individual's immune system producing its own antibodies and memory cells in response to an antigen, which takes time to develop but is long-lasting. Passive immunity involves the direct introduction of external antibodies, providing immediate but short-term protection without memory cell production.
評分準則
1 mark for stating active immunity involves memory cells/antibody production by the host, whereas passive does not. 1.3 marks for stating active immunity is long-term/slow, whereas passive is immediate/short-term.
題目 12 · short-answer
2.3 分
An electron micrograph of a mitochondrion has a measured length of 48 mm. The magnification of the image is x30,000. Calculate the actual length of the mitochondrion in micrometres (microns). Show your working.
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解題
Actual size = Image size / Magnification. Image size = 48 mm = 48,000 micrometres. Actual size = 48,000 / 30,000 = 1.6 micrometres.
評分準則
1 mark for converting mm to micrometres correctly (48,000) or setting up the correct equation (48 / 30,000). 1.3 marks for the correct final answer of 1.6 micrometres.
題目 13 · short-answer
2.3 分
In a sample of root tip cells undergoing mitosis, 12 cells were in prophase, 5 in metaphase, 3 in anaphase, 4 in telophase, and 216 in interphase. Calculate the mitotic index of this cell sample. Give your answer to one decimal place.
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解題
Total number of cells = 12 + 5 + 3 + 4 + 216 = 240 cells. Number of cells in mitosis = 12 + 5 + 3 + 4 = 24 cells. Mitotic index = (24 / 240) * 100 = 10.0%.
評分準則
1 mark for calculating the total number of cells (240) and mitotic cells (24). 1.3 marks for the correct percentage of 10.0% (accept 10%).
題目 14 · short-answer
2.3 分
Define the term 'species' and explain how phylogenetic classification differs from simple artificial classification.
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解題
A species consists of organisms with similar morphological, physiological, and behavioural features that can interbreed to produce fertile offspring. Phylogenetic classification groups organisms based on evolutionary history, shared ancestors, and genetic relationships, whereas artificial classification groups organisms solely by non-evolutionary, easily observable physical features.
評分準則
1 mark for defining species as organisms that breed to produce fertile offspring. 1.3 marks for explaining that phylogenetic classification is based on evolutionary relationships/ancestry.
題目 15 · short-answer
2.3 分
Explain the role of hydrogen bonds in the cohesion-tension theory of water transport in xylem vessels.
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解題
Water molecules are polar and form hydrogen bonds with one another. This attraction is called cohesion. When water evaporates from the leaves (transpiration), it pulls the next water molecule due to cohesion, creating tension. This allows a continuous, unbroken column of water to be drawn up the xylem from the roots.
評分準則
1 mark for identifying that hydrogen bonding causes cohesion/attraction between water molecules. 1.3 marks for explaining this creates a continuous column of water pulled up under tension.
題目 16 · short-answer
2.3 分
Explain how the active transport of sodium ions out of epithelial cells into the blood allows the co-transport of glucose from the lumen of the ileum.
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解題
Sodium-potassium pumps actively transport sodium ions out of the epithelial cells into the blood. This lowers the concentration of sodium ions inside the cytoplasm of the cell compared to the lumen. Sodium ions then diffuse down their concentration gradient into the cell through a co-transporter protein, carrying glucose molecules with them against their concentration gradient.
評分準則
1 mark for explaining that active transport lowers intracellular sodium ion concentration, creating a concentration gradient between the lumen and the cell. 1.3 marks for explaining that sodium ions enter the cell by facilitated diffusion via a co-transporter protein, pulling glucose along with them.
題目 17 · Recall & Short Structured
2 分
At the arteriole end of a capillary bed, the hydrostatic pressure of the blood is \(4.6\text{ kPa}\) and the hydrostatic pressure of the surrounding tissue fluid is \(1.3\text{ kPa}\). The water potential gradient is maintained by an oncotic pressure of \(3.1\text{ kPa}\) in the blood and \(0.9\text{ kPa}\) in the tissue fluid. Calculate the net filtration pressure directing fluid out of the capillary. Show your working.
1 mark for showing correct working: \( (4.6 - 1.3) - (3.1 - 0.9) \) or \( 3.3 - 2.2 \). 1 mark for correct final answer of \( 1.1\text{ kPa} \) (accept without units, but reject if incorrect units are given).
題目 18 · Recall & Short Structured
3 分
During periods of intense flight, the muscle cells of a locust respire anaerobically and produce lactate. Explain how the production of lactate leads to an increased rate of oxygen diffusion into these active muscle tissues.
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解題
Lactate lowers the water potential of the muscle cells. Consequently, water moves from the fluid-filled tracheoles into the muscle cells via osmosis. This withdrawal of water from the tracheoles increases the surface area of the tracheoles exposed to the air and shortens the diffusion path for oxygen gas, allowing it to diffuse much faster.
評分準則
1 mark for stating lactate lowers the water potential of the muscle cells. 1 mark for stating water moves from the tracheoles into the muscle cells by osmosis. 1 mark for explaining that this shortens the diffusion pathway (or increases direct contact of air with muscle cells), increasing the rate of oxygen diffusion.
題目 19 · Recall & Short Structured
3 分
When preparing a sample of plant cells to isolate chloroplasts via cell fractionation, the tissue must be placed in a solution that is cold, isotonic, and buffered. Explain the importance of each of these three conditions.
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解題
1. Cold: reduces kinetic energy of hydrolytic enzymes that could break down organelles. 2. Isotonic: maintains same water potential to prevent osmotic movement of water, avoiding lysis or shrinking of chloroplasts. 3. Buffered: maintains constant pH to prevent the denaturation of enzymes and membrane proteins.
評分準則
1 mark for: Cold reduces enzyme activity to prevent self-digestion/breakdown of organelles. 1 mark for: Isotonic prevents net movement of water by osmosis, preventing organelle lysis/shrinking. 1 mark for: Buffered maintains a constant pH to prevent proteins/enzymes from denaturing.
題目 20 · Recall & Short Structured
2 分
Biological classification systems use a phylogenetic hierarchy. Explain what is meant by a phylogenetic hierarchy.
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解題
A phylogenetic hierarchy is a classification system where organisms are grouped according to their shared evolutionary history and common ancestors. It is a hierarchy because smaller groups are nested inside larger groups, and there is no overlap between different groups at the same taxonomic level.
評分準則
1 mark for: groups are nested/placed within larger groups with no overlap. 1 mark for: classification is based on evolutionary relationships or shared common ancestry.
題目 21 · Extended Response
3.625 分
A student prepared a temporary mount of a transverse section of a herbaceous stem to observe xylem vessels under a light microscope. Explain how the student should use a stage micrometer and an eyepiece graticule to measure the actual mean lumen diameter of the xylem vessels.
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解題
1. Place the stage micrometer on the microscope stage and align its scale with the eyepiece graticule. 2. Count the number of eyepiece graticule units (egus) that correspond to a known distance on the stage micrometer. 3. Calculate the calibration factor (the distance in micrometres represented by one egu, e.g., if \(40\text{ egus} = 100\,\mu\text{m}\), then \(1\text{ egu} = 2.5\,\mu\text{m}\)). 4. Replace the stage micrometer with the stem specimen slide, measure the diameter of several xylem vessel lumens using the graticule, convert these to micrometres, and calculate the mean value.
評分準則
1 mark for aligning stage micrometer with eyepiece graticule and counting overlapping divisions. 1 mark for showing the calculation of the value of one eyepiece division (calibration factor). 1 mark for measuring several xylem vessel diameters in egus and converting to real length (\(\mu\text{m}\)) using the calibration factor. 0.625 marks for describing how to calculate the mean diameter from multiple measurements.
題目 22 · Extended Response
3.625 分
An investigation was carried out to measure the rate of transpiration of a leafy shoot using a potometer. During the assembly, the student made sure to cut the shoot under water and to seal all joints with petroleum jelly. Explain the biological and practical reasons for these two steps.
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解題
Cutting the shoot under water prevents air bubbles from entering the cut xylem vessels. If air enters, it breaks the continuous water column and disrupts the cohesion-tension mechanism, stopping transpiration pull. Sealing the joints with petroleum jelly ensures the apparatus is completely airtight. This prevents air leaks which would stop the meniscus/bubble from moving and ensure that any water lost by transpiration is accurately reflected by water uptake measured by the bubble.
評分準則
1.625 marks for explaining cutting under water (0.8125 marks for preventing air entry into xylem, 0.8125 marks for maintaining the continuous water column / cohesive forces / transpiration stream). 2 marks for explaining sealing with petroleum jelly (1 mark for ensuring the apparatus is airtight / preventing external leaks, 1 mark for ensuring bubble movement accurately reflects water uptake / transpiration rate).
題目 23 · Extended Response
3.625 分
Describe the structural adaptations of the gas exchange system of an insect that allow efficient oxygen delivery to respiring tissues while minimizing water loss, and explain how the movement of abdominal muscles aids in this process.
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解題
Insects have branched tracheae and tracheoles which provide a large surface area and short diffusion path directly to cells. To minimize water loss, they have a waterproof chitinous cuticle and spiracles controlled by valves that can close. When active, fluid at the ends of the tracheoles is drawn into tissues by osmosis, speeding up diffusion of gases. Contraction of abdominal muscles squeezes the trachea, causing mass flow of air in and out along a pressure gradient (ventilation), maintaining a steep concentration gradient for rapid gas exchange.
評分準則
1 mark for tracheae and highly branched tracheoles delivering oxygen directly to tissues (and/or fluid moving out of tracheoles during activity to shorten diffusion distance). 1 mark for waterproof exoskeleton/cuticle and spiracles with valves that close to reduce water loss. 1 mark for abdominal contraction causing mass flow of air / ventilation due to pressure changes. 0.625 marks for explaining that ventilation maintains a steep concentration gradient for rapid gas exchange during high respiratory demand.
題目 24 · Extended Response
3.625 分
A student drew a scientific diagram of a single gill lamella under high power. Describe the rules the student must follow when producing a high-power scientific drawing of this anatomical structure from a slide.
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解題
When drawing a scientific diagram of a gill lamella, the student must: 1. Use a sharp HB pencil to draw single, clear, continuous lines with no sketching or shading. 2. Ensure all structures are drawn in correct proportion and reflect what is actually seen. 3. Include a suitable title, scale bar, or magnification. 4. Use straight, horizontal, non-overlapping label lines that touch the target structure directly without arrowheads.
評分準則
1 mark for using a sharp pencil and making single, continuous lines with no sketching or shading. 1 mark for ensuring accurate proportions and representing only what is visible on the slide. 1 mark for drawing straight, non-crossing label lines that touch the structures without arrowheads. 0.625 marks for including a clear title, magnification, or a scale bar.
題目 25 · Extended Response
3.625 分
Monoclonal antibodies can be conjugated to therapeutic drugs to target cancer cells specifically. Describe how monoclonal antibodies target cancer cells and explain the advantage of using conjugated monoclonal antibodies compared to non-selective chemotherapy.
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解題
Monoclonal antibodies have a highly specific tertiary structure with variable regions (antigen-binding sites) that are complementary only to specific tumor-associated antigens found on cancer cells. They bind to these antigens to form antigen-antibody complexes, delivering the conjugated therapeutic drug directly to the cancer cells. This is advantageous because non-selective chemotherapy damages healthy, rapidly-dividing cells, whereas conjugated antibodies spare healthy body cells (as they lack the target antigen), reducing side effects and requiring a lower overall dosage.
評分準則
1 mark for describing the complementary shape of the antibody's variable region/tertiary structure to the tumor-associated antigen. 1 mark for explaining that the antibody binds specifically to cancer cells to form antigen-antibody complexes. 1 mark for stating that the conjugated drug is delivered directly to cancer cells, sparing healthy cells. 0.625 marks for explaining that this reduces side effects/toxicity or allows a lower effective therapeutic dose.
題目 26 · Extended Response
3.625 分
An ultracentrifuge was used to isolate organelles from a sample of homogenized plant tissue. Explain why the tissue was kept in a cold, isotonic, and buffered solution before and during homogenization, and list the order in which nuclei, chloroplasts, and mitochondria would sediment out.
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解題
The solution is kept: 1. Cold to reduce enzyme activity, preventing self-digestion/autolysis of organelles. 2. Isotonic to prevent osmotic water movement, ensuring organelles do not burst (lysis) or shrink (crenation). 3. Buffered to maintain a constant pH, preventing denaturation of organelle proteins and enzymes. When centrifuged at progressively higher speeds, organelles sediment out in order of size/density: nuclei first (at low speed), chloroplasts second, and mitochondria third.
評分準則
1.625 marks for the functions of the solution (0.5417 marks each: cold reduces enzyme activity to prevent damage; isotonic prevents water movement via osmosis to stop organelles bursting/shrinking; buffered maintains constant pH to prevent protein denaturation). 2 marks for listing the correct order of sedimentation (1 mark for nuclei first, 1 mark for chloroplasts second, and mitochondria third; accept chloroplasts then mitochondria based on chloroplasts being larger than mitochondria).
題目 27 · Extended Response
3.625 分
Describe the process of binary fission in prokaryotic cells and explain how it differs from eukaryotic mitosis.
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解題
In binary fission, the circular DNA molecule replicates once, while plasmids can replicate multiple times. The cell elongates and the two copies of circular DNA move to opposite poles of the cell. The cytoplasm then divides (cytokinesis) and a new cell wall develops, producing two genetically identical daughter cells. This differs from mitosis because prokaryotes do not undergo nuclear envelope breakdown (as they lack a nucleus), do not condense chromosomes, and do not use a spindle apparatus to separate genetic material.
評分準則
1 mark for replication of circular DNA (once) and plasmids (variable times). 1 mark for elongation of the cell and migration of DNA loops to opposite poles, followed by cytoplasm division / new cell wall formation. 1 mark for stating key differences from mitosis (no nuclear envelope breakdown, no visible chromosome condensation, no spindle fibers/apparatus). 0.625 marks for noting that plasmids are distributed randomly/unequally between daughter cells, unlike chromosomes in mitosis.
題目 28 · Extended Response
3.625 分
Explain how the comparison of amino acid sequences of a highly conserved protein, such as cytochrome c, can be used to determine the evolutionary relationships between different species of primates.
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解題
Cytochrome c is a highly conserved protein found in all primates because it is essential for respiration. By sequencing cytochrome c from different primates, scientists can compare the order of amino acids and count the number of differences. A smaller number of amino acid differences indicates a closer evolutionary relationship and a more recent common ancestor, because there has been less time for mutations to accumulate in the gene coding for the protein. A larger number of differences indicates a more distant common ancestor.
評分準則
1 mark for comparing the sequence/order of amino acids in the cytochrome c protein between species. 1 mark for explaining that fewer differences in the sequence indicate a more recent common ancestor / closer evolutionary relationship. 1 mark for explaining that more differences indicate a more distant common ancestor / more time since divergence. 0.625 marks for explaining that differences in the amino acid sequence are caused by mutations accumulating in the DNA/gene over time.
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