An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V2) Cambridge International A Level Human Biology paper. Not affiliated with or reproduced from Cambridge.
Paper 1R
Answer ALL questions. Show all the steps in any calculations and state the units. Use of a calculator is permitted.
26 題目 · 77 分
題目 1 · Sentence Completion
8 分
Complete the following passage about the function of the nephron. Write a suitable word or words in each of the blank spaces. The process of ultrafiltration occurs in the Bowman's capsule of the nephron. Blood enters the glomerulus via the (1) ________ arteriole, which has a wider lumen than the (2) ________ arteriole. This difference in diameter creates a high (3) ________ pressure, forcing small molecules out of the blood. The liquid that enters the nephron is called the glomerular (4) ________. This fluid contains glucose, which is completely reabsorbed back into the blood at the (5) ________ convoluted tubule. This reabsorption occurs via the process of (6) ________ transport, which requires energy in the form of (7) ________ produced during aerobic respiration. Most of the water is reabsorbed further down the nephron, particularly under the influence of the hormone (8) ________.
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解題
1: The afferent arteriole is the wider blood vessel bringing blood into the glomerulus. 2: The efferent arteriole is narrower, leaving the glomerulus. 3: This bottleneck creates a high hydrostatic pressure that drives ultrafiltration. 4: The fluid containing water, salts, glucose, and urea is the glomerular filtrate. 5: Glucose is selectively reabsorbed at the proximal convoluted tubule. 6: This selective reabsorption requires active transport. 7: Active transport is powered by ATP. 8: Water reabsorption in the collecting duct is regulated by ADH (antidiuretic hormone).
評分準則
Award 1 mark for each correct term: (1) afferent [reject: efferent]; (2) efferent [reject: afferent]; (3) hydrostatic / blood [accept: high]; (4) filtrate; (5) proximal; (6) active; (7) ATP [accept: adenosine triphosphate]; (8) ADH / antidiuretic hormone.
題目 2 · Calculation
3 分
A student measures the diameter of a human red blood cell in a micrograph. The diameter of the cell on the micrograph is measured to be \(3.5\text{ cm}\). The actual diameter of this red blood cell is known to be \(7\ \mu\text{m}\). Calculate the magnification of the micrograph. Show your working.
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解題
To calculate magnification, use the formula: \[\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\]
First, convert both measurements to the same units. Converting centimeters to micrometers (\(\mu\text{m}\)): \[3.5\text{ cm} = 35\text{ mm}\] \[35\text{ mm} \times 1000 = 35,000\ \mu\text{m}\]
Now, divide the image size by the actual size: \[\text{Magnification} = \frac{35,000\ \mu\text{m}}{7\ \mu\text{m}} = 5000\]
Therefore, the magnification of the micrograph is \(\times 5000\).
評分準則
Award marks as follows: - **MP1 (Unit Conversion):** Convert \(3.5\text{ cm}\) to \(35,000\ \mu\text{m}\) OR convert \(7\ \mu\text{m}\) to \(0.0007\text{ cm}\) (1 mark) - **MP2 (Method):** Divide image size by actual size, e.g., \(\frac{35,000}{7}\) or \(\frac{3.5}{0.0007}\) (1 mark) - **MP3 (Accuracy):** Correct evaluation of magnification as \(5000\) or \(\times 5000\) (1 mark)
*Note: Allow full marks for a correct final answer with no working shown.*
題目 3 · Calculation
3 分
A person has a stroke volume of \(75\text{ cm}^3\). Their heart rate is measured by counting \(21\) heartbeats over a period of \(15\text{ seconds}\). Calculate the cardiac output of this person in \(dm^3 / min\). Show your working.
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解題
First, calculate the heart rate in beats per minute (bpm): \[21\text{ beats in } 15\text{ seconds} \implies 21 \times 4 = 84\text{ beats per minute}\]
Next, use the formula for cardiac output: \[\text{Cardiac Output} = \text{Stroke Volume} \times \text{Heart Rate}\] \[\text{Cardiac Output} = 75\text{ cm}^3 \times 84\text{ bpm} = 6300\text{ cm}^3 / min\]
Finally, convert the cardiac output from \(\text{cm}^3\) to \(\text{dm}^3\) (since \(1\text{ dm}^3 = 1000\text{ cm}^3\)): \[\frac{6300\text{ cm}^3}{1000} = 6.3\text{ dm}^3 / min\]
評分準則
Award marks as follows: - **MP1 (Heart Rate):** Calculate heart rate per minute: \(21 \times 4 = 84\text{ bpm}\) (1 mark) - **MP2 (Cardiac Output in \(cm^3\)):** Multiply stroke volume by heart rate: \(75 \times 84 = 6300\text{ cm}^3 / min\) (1 mark) - **MP3 (Accuracy & Conversion):** Divide by \(1000\) to give a final answer of \(6.3\) (1 mark)
*Note: Allow full marks for a correct final answer with no working shown.*
題目 4 · Diagram Drawing / Labeling
3 分
The diagram below describes a typical motor neurone with three labeled structures, X, Y, and Z. - Structure X represents the branched projections that receive incoming signals from other neurones. - Structure Y represents the long, single fiber that conducts nerve impulses away from the cell body. - Structure Z represents the fatty insulating layer surrounding the long fiber.
Identify structures X, Y, and Z.
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解題
Structure X represents the dendrites (or dendrons) which function to receive nerve impulses from other neurones and carry them towards the cell body. Structure Y is the axon, which is a long cytoplasmic extension that transmits electrical impulses away from the cell body towards the axon terminals. Structure Z is the myelin sheath, which is formed by Schwann cells and acts as an electrical insulator to speed up the transmission of nerve impulses.
評分準則
Award 1 mark for each correct identification: - X: Dendrite / Dendron (1) - Y: Axon (1) - Z: Myelin sheath (1)
Do not accept: myelin alone for Z; cell body for X.
題目 5 · Diagram Drawing / Labeling
3 分
A student is studying a diagram of a single villus from the human small intestine. The diagram has three labeled components: - Label A points to the single-cell thick outer layer that absorbs nutrients. - Label B points to the central lymphatic vessel that absorbs fatty acids and glycerol. - Label C points to the network of tiny blood vessels that transport glucose and amino acids.
State the names of structures A, B, and C.
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解題
Structure A is the epithelium (or epithelial cell layer), which is only one cell thick to minimize the diffusion distance for digested nutrients. Structure B is the lacteal, a specialized lymphatic vessel located in the center of the villus that absorbs and transports lipids (fatty acids and glycerol). Structure C represents the capillary network (or blood capillaries), which rapidly transport absorbed water-soluble molecules like glucose, amino acids, minerals, and water away from the intestine to maintain a steep concentration gradient.
評分準則
Award 1 mark for each correct identification: - A: Epithelium / Epithelial cell(s) (1) - B: Lacteal (1) - C: Capillary / Blood capillary / Capillary network (1)
Do not accept: lymph vessel for B; blood vessel (too general) for C.
題目 6 · Diagram Drawing / Labeling
3 分
In a diagram of a synovial joint, such as the human knee joint, three functional parts are highlighted: - Part P is a smooth, slippery tissue covering the ends of the bones to reduce friction and absorb shock during movement. - Part Q is the fluid-filled space that lubricates the joint cavity. - Part R is the tough, fibrous tissue that connects bone to bone, holding the joint stable.
Name the structures P, Q, and R.
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解題
Part P is the cartilage (specifically articular cartilage) which prevents bone-to-bone contact and absorbs impact. Part Q is the synovial fluid, which is secreted by the synovial membrane to lubricate the joint and reduce friction during movement. Part R is the ligament, a strong band of fibrous connective tissue that connects bones together and prevents dislocation.
評分準則
Award 1 mark for each correct identification: - P: Cartilage / Articular cartilage (1) - Q: Synovial fluid (1) - R: Ligament (1)
Do not accept: tendon for R; synovial membrane for Q.
題目 7 · Diagram Drawing / Labeling
3 分
A diagram showing a vertical section of human skin contains three main structures involved in regulating body temperature: - Structure J is a coiled tubular gland that secretes a liquid onto the skin surface to cool the body. - Structure K is a tiny muscle attached to a hair follicle that contracts to raise the hair. - Structure L is a blood vessel near the skin surface that can widen (dilate) to increase heat loss.
Identify structures J, K, and L.
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解題
Structure J is the sweat gland, which extracts water, salts, and urea from blood to produce sweat for evaporative cooling. Structure K is the erector pili muscle (or hair erector muscle), which contracts in response to cold to pull the hair upright, trapping an insulating layer of warm air. Structure L is an arteriole (or capillary / subcutaneous blood vessel) that dilates (vasodilation) during overheating to allow more blood to flow close to the skin surface, increasing heat loss by radiation.
評分準則
Award 1 mark for each correct identification: - J: Sweat gland (1) - K: Erector pili muscle / Hair erector muscle (1) - L: Arteriole / Capillary / Blood vessel / Shunt vessel (1)
Do not accept: sweat duct for J; vein or artery for L.
題目 8 · Diagram Drawing / Labeling
3 分
A diagram shows the interface where gas exchange occurs between an alveolus and a surrounding blood capillary in the human lungs. - Structure D is the thin layer of moisture lining the inside of the alveolus. - Structure E is the ultra-thin cell layer forming the wall of the alveolus. - Structure F is the biconcave cell inside the capillary that transports oxygen.
Give the names of structures D, E, and F.
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解題
Structure D is the moisture layer (containing surfactant) which dissolves oxygen before it diffuses across the respiratory membrane and prevents alveolar collapse. Structure E is the squamous epithelium (or alveolar wall), which consists of extremely thin, flat cells to minimize the diffusion pathway for carbon dioxide and oxygen. Structure F is the red blood cell (or erythrocyte), which contains hemoglobin to bind and transport oxygen to tissues.
評分準則
Award 1 mark for each correct identification: - D: Moisture layer / Surfactant / Liquid film (1) - E: Alveolar epithelium / Squamous epithelium / Alveolar wall / Epithelial cell (1) - F: Red blood cell / Erythrocyte (1)
Do not accept: cell wall for E; white blood cell for F.
題目 9 · Short Answer
2 分
State two ways in which the structure of a virus differs from the structure of a bacterium.
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解題
Viruses are acellular and lack cytoplasm, organelles, or cell membranes, unlike bacteria. Viruses also possess a simple protein coat (capsid) surrounding their genetic material (which can be DNA or RNA), whereas bacteria have a cell wall made of peptidoglycan and circular DNA.
評分準則
1 mark for identifying that viruses lack a cellular structure (or cytoplasm, organelles) whereas bacteria are cellular. 1 mark for stating that viruses have a protein coat (capsid) surrounding genetic material, whereas bacteria have a cell wall / cell membrane / circular DNA.
題目 10 · Short Answer
2 分
A student conducts an experiment to find the energy content of a food sample. Burning a \(1.5\text{ g}\) sample of food raises the temperature of \(20\text{ g}\) of water by \(18^\circ\text{C}\). Calculate the energy released per gram of food. (Specific heat capacity of water is \(4.2\text{ J/g}^\circ\text{C}\)).
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解題
Energy released = \(\text{mass of water} \times \text{specific heat capacity} \times \text{temperature rise}\). Energy = \(20\text{ g} \times 4.2\text{ J/g}^\circ\text{C} \times 18^\circ\text{C} = 1512\text{ J}\). Energy per gram of food = \(\frac{1512\text{ J}}{1.5\text{ g}} = 1008\text{ J/g}\).
評分準則
1 mark for calculating the total energy released as \(1512\text{ J}\) (or \(1.512\text{ kJ}\)). 1 mark for dividing by the mass of the food to obtain \(1008\text{ J/g}\) (or \(1.008\text{ kJ/g}\)). Allow 1 mark for correct calculation step with a minor arithmetic error.
題目 11 · Short Answer
2 分
State the primary function of ciliated epithelial cells in the human respiratory tract and describe how their structure helps them perform this function.
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解題
Ciliated epithelial cells possess tiny, hair-like projections called cilia on their free surface. These cilia beat in a synchronized, wave-like pattern to move mucus (which traps dust and pathogens) upwards and outwards from the airways, protecting the lungs.
評分準則
1 mark for stating the function: moving/sweeping mucus (and trapped particles/pathogens) away from the lungs / towards the throat. 1 mark for explaining the structural adaptation: possession of hair-like cilia that beat/flick.
題目 12 · Short Answer
2 分
Describe the chemical test used to confirm the presence of non-reducing sugars, such as sucrose, in a food solution.
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解題
To test for a non-reducing sugar, first heat the sample with dilute hydrochloric acid to hydrolyse the glycosidic bonds, breaking sucrose into glucose and fructose. Next, neutralise the acid by adding sodium hydrogencarbonate. Finally, add Benedict's reagent and heat the mixture in a hot water bath. A change from blue to green, orange, or brick-red indicates the presence of non-reducing sugars.
評分準則
1 mark for heating the sample with acid (to hydrolyse) and neutralising with alkali/sodium hydrogencarbonate. 1 mark for adding Benedict's reagent and heating, observing a color change (blue to green/yellow/orange/red).
題目 13 · Short Answer
2 分
Identify the type of joint found at the elbow and describe the range of movement it permits compared to the joint at the shoulder.
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解題
The elbow is an example of a hinge joint, which allows movement back and forth in a single plane (flexion and extension). In contrast, the shoulder is a ball-and-socket joint, allowing a much wider, 360-degree rotational range of movement in all three planes.
評分準則
1 mark for identifying the elbow as a hinge joint. 1 mark for comparing the movement: elbow permits movement in one plane / direction (flexion/extension), while the shoulder (ball-and-socket) allows movement in multiple planes / rotational movement.
題目 14 · Short Answer
2 分
Explain how the body responds to a decrease in core body temperature to reduce heat loss through the skin.
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解題
When the core body temperature drops, vasoconstriction occurs. Arterioles supplying the skin capillaries constrict, reducing the volume of warm blood flowing close to the skin surface, thereby reducing heat loss by radiation. Additionally, hair erector muscles contract (causing goosebumps) to erect hairs and trap an insulating layer of warm air next to the skin.
評分準則
1 mark for describing vasoconstriction (narrowing of arterioles supplying skin capillaries). 1 mark for explaining that this reduces blood flow to the skin surface, thereby reducing heat loss by radiation/convection. (Alternatively, accept hair erector muscles contracting to trap an insulating layer of air).
題目 15 · Short Answer
2 分
Describe how a nerve impulse is transmitted across a synapse from one neurone to the next.
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解題
When an electrical impulse reaches the pre-synaptic membrane, it stimulates vesicles containing neurotransmitters to fuse with the membrane, releasing them into the synaptic cleft via exocytosis. The neurotransmitters diffuse across the synaptic gap and bind to specific receptors on the post-synaptic membrane, initiating a new electrical impulse in the next neurone.
評分準則
1 mark for mentioning the release and diffusion of neurotransmitters across the synaptic gap/cleft. 1 mark for stating that neurotransmitters bind to receptors on the post-synaptic membrane to trigger a new impulse.
題目 16 · Short Answer
2 分
State two structural differences between an artery and a vein.
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解題
Arteries have thick walls containing high amounts of muscle and elastic tissue to withstand high pressure, and they have a narrow lumen. Veins have much thinner walls with less muscle and elastic tissue, a wider lumen, and contain valves to prevent the backflow of slow-moving blood.
評分準則
1 mark for mentioning the difference in wall thickness / muscle and elastic fibers (arteries thicker, veins thinner) OR difference in lumen size (arteries narrow, veins wide). 1 mark for mentioning that veins have valves while arteries do not (except semilunar valves at the exit of the heart).
題目 17 · short_answer
2 分
State two differences between active transport and facilitated diffusion.
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解題
1. Active transport is an active process that requires metabolic energy in the form of ATP, whereas facilitated diffusion is a passive process that relies only on the kinetic energy of particles. 2. Active transport moves substances against a concentration gradient (from a region of lower concentration to a region of higher concentration), while facilitated diffusion moves substances down a concentration gradient (from a region of higher concentration to a region of lower concentration).
評分準則
Award 1 mark for each correct difference up to a maximum of 2 marks: - Active transport requires energy/ATP whereas facilitated diffusion does not/is passive (1 mark). - Active transport moves substances against/up the concentration gradient whereas facilitated diffusion moves substances down the concentration gradient (1 mark).
題目 18 · short_answer
2 分
Describe two structural features of capillaries that adapt them for the efficient exchange of substances with body cells.
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解題
Capillaries have walls composed of a single layer of squamous endothelial cells (one cell thick), minimizing the diffusion distance for gases and nutrients. Additionally, they have tiny gaps or pores between the endothelial cells that increase permeability to water and dissolved solutes.
評分準則
Award 1 mark for each valid structural feature linked to its adaptation (maximum 2 marks): - Wall is one cell thick / thin wall (1) which provides a short diffusion distance (1). - Narrow lumen (1) which slows down blood flow / pushes red blood cells close to the capillary wall for efficient diffusion (1). - Gaps/pores between endothelial cells (1) which increases permeability to water and small solutes (1).
題目 19 · short_answer
2 分
Describe how a student would test a food sample to confirm the presence of lipid (fat) using the emulsion test.
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解題
To perform the emulsion test, first mix the food sample thoroughly with ethanol and shake to dissolve any lipids. Filter or decant the liquid into another test tube containing clean water. A positive result is indicated by the formation of a cloudy white emulsion.
評分準則
Award 1 mark for the method and 1 mark for the positive result: - Add ethanol to the sample, shake/mix, and then add this mixture to water (1 mark). - Observe the formation of a cloudy-white emulsion/suspension (1 mark). [Reject: 'precipitate' for emulsion, reject 'add water then ethanol' without mixing first]
題目 20 · short_answer
2 分
Explain why penicillin is highly effective at treating a bacterial infection like throat inflammation but has no effect on a common viral cold.
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解題
Penicillin is an antibiotic that specifically inhibits the synthesis of peptidoglycan, a crucial component of bacterial cell walls, causing osmotic lysis of the bacteria. Viruses do not possess cell walls, cytoplasm, or their own metabolic machinery; instead, they hijack host cellular machinery to replicate, making them completely unaffected by drugs targeting bacterial structures.
評分準則
Award 1 mark for explaining why it works on bacteria and 1 mark for explaining why it fails on viruses: - Penicillin inhibits bacterial cell wall/peptidoglycan synthesis, causing the bacteria to burst/die (1 mark). - Viruses do not have cell walls/metabolic machinery / viruses are non-cellular (1 mark). [Accept: 'viruses live inside host cells where antibiotics cannot target them' as an alternative second point]
題目 21 · Extended Answer
4 分
Explain how vaccination protects an individual against a specific bacterial disease.
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解題
Vaccines contain dead or weakened forms of the pathogen, or its isolated antigens, which are injected into the body. These antigens are recognized by the immune system, stimulating specific B-lymphocytes to divide and produce antibodies. Some of these lymphocytes develop into long-lived memory cells. If the individual is later exposed to the same live pathogen, these memory cells recognize the antigens immediately, leading to a much faster, rapid, and larger production of antibodies, which neutralizes or destroys the pathogen before it can cause disease symptoms.
評分準則
1. Vaccine contains dead/attenuated/weakened pathogens or antigens (1 mark) 2. Antigens trigger an immune response where lymphocytes produce specific antibodies (1 mark) 3. Memory cells are produced and remain in the body (1 mark) 4. Upon future exposure, memory cells produce antibodies much faster and in larger quantities to destroy the bacteria (1 mark)
題目 22 · Extended Answer
4 分
Explain the roles of bile in the digestion and absorption of lipids in the human digestive system.
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解題
Bile is an alkaline fluid produced by the liver and stored in the gallbladder. When it enters the small intestine, it neutralises the acidic chyme coming from the stomach, which creates the optimal alkaline environment for digestive enzymes like lipase to function. Furthermore, bile emulsifies lipids, meaning it breaks down large fat globules into many tiny droplets. This physical process significantly increases the overall surface area of the lipids available for pancreatic lipase to chemically break down into fatty acids and glycerol, leading to more efficient absorption.
評分準則
1. Neutralises acidic chyme / provides alkaline conditions (1 mark) 2. Provides optimum pH for lipase activity (1 mark) 3. Emulsifies lipids / breaks large fat globules into smaller droplets (1 mark) 4. Increases surface area for lipase to act upon, speeding up digestion (1 mark)
題目 23 · Extended Answer
4 分
Explain how the structure of ciliated epithelial cells in the trachea is adapted to protect the lungs from pathogens and dust.
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解題
Ciliated epithelial cells line the trachea and possess specialized, hair-like structures on their surface called cilia. Goblet cells interspersed among them secrete sticky mucus, which traps inhaled dust particles, bacteria, and other pathogens. The cilia undergo coordinated, rhythmic beating movements to push this trapped mucus upwards, away from the respiratory tract and towards the pharynx (throat). From there, the mucus can be swallowed and destroyed by stomach acid, or coughed out, thereby preventing infection and blockages in the lungs.
評分準則
1. Goblet cells secrete mucus to trap dust and pathogens (1 mark) 2. Cilia are tiny hair-like projections on the cell surface (1 mark) 3. Cilia beat in a coordinated, rhythmic wave (1 mark) 4. Mucus is moved upwards, away from the lungs towards the throat to be swallowed/coughed out (1 mark)
題目 24 · Extended Answer
4 分
Explain the homeostatic response of the blood vessels in the skin when a person is exposed to a very cold environment.
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解題
When exposed to a cold environment, thermoreceptors detect the decrease in temperature and send signals to the hypothalamus. In response, vasoconstriction occurs. The muscles in the walls of the arterioles supplying the superficial capillary loops near the skin surface contract, narrowing the arteriole lumen. This significantly reduces the volume of warm blood flowing close to the skin surface. Consequently, less thermal energy (heat) is lost from the blood to the external surroundings via radiation, convection, and conduction, helping to maintain core body temperature.
評分準則
1. Vasoconstriction occurs in response to cold (1 mark) 2. Arterioles supplying superficial capillaries constrict/narrow (1 mark) 3. Blood flow close to the skin surface is reduced (1 mark) 4. Less heat is lost to the external environment via radiation/conduction/convection (1 mark) (Reject: capillaries constrict)
題目 25 · Extended Answer
4 分
Explain how active transport differs from facilitated diffusion in terms of concentration gradients, energy requirements, and the membrane components involved.
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解題
Active transport and facilitated diffusion are two distinct mechanisms for moving substances across cell membranes. Firstly, active transport moves molecules or ions against their concentration gradient (from a region of low concentration to a region of high concentration), whereas facilitated diffusion moves substances down their concentration gradient (from high to low concentration). Secondly, active transport is an active process requiring metabolic energy in the form of ATP (produced by aerobic respiration), while facilitated diffusion is entirely passive and does not require energy. Finally, active transport relies specifically on carrier proteins (pumps) to transport molecules, whereas facilitated diffusion can utilize both channel proteins and carrier proteins to assist the movement.
評分準則
1. Active transport is against the concentration gradient (low to high), facilitated diffusion is down/along the concentration gradient (high to low) (1 mark) 2. Active transport requires energy/ATP, facilitated diffusion is passive/does not require energy (1 mark) 3. Active transport uses carrier proteins only (1 mark) 4. Facilitated diffusion uses both channel and carrier proteins (1 mark)
題目 26 · Extended Answer
4 分
Explain how the structure of a capillary is adapted to its function of exchanging substances between the blood and tissue fluid.
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解題
Capillaries are highly specialized vessels designed for the efficient exchange of gases, nutrients, and waste products between blood and body tissues. Their walls are extremely thin, consisting of only a single layer of squamous endothelial cells, which provides a very short diffusion pathway for substances like oxygen, glucose, and carbon dioxide. Furthermore, the capillary walls contain microscopic pores or intercellular clefts, allowing water, ions, and small soluble molecules to pass through easily. Lastly, capillaries have a very narrow lumen (approximately the width of a single red blood cell), which forces red blood cells to travel in single file. This slows down blood flow, maximizing the time available for diffusion to occur, while ensuring the cells are as close as possible to the surrounding tissue.
評分準則
1. Wall is one cell thick / made of thin endothelial cells (1 mark) 2. Provides a very short diffusion distance/pathway (1 mark) 3. Pores/clefts between cells allow easy passage of small molecules/water (1 mark) 4. Narrow lumen slows blood flow / brings red blood cells close to capillary wall to maximize exchange (1 mark)
Paper 2R
Answer ALL questions. Show all the steps in any calculations and state the units. Use of a calculator is permitted.
29 題目 · 80 分
題目 1 · 選擇題
1 分
A person drinks a large volume of water. Which of the following correctly describes the subsequent changes in the concentration of anti-diuretic hormone (ADH) in the blood and the permeability of the collecting ducts in the kidneys?
A.ADH concentration increases, and the permeability of the collecting ducts increases
B.ADH concentration increases, and the permeability of the collecting ducts decreases
C.ADH concentration decreases, and the permeability of the collecting ducts increases
D.ADH concentration decreases, and the permeability of the collecting ducts decreases
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解題
When a person drinks a large volume of water, the blood plasma becomes more dilute. This change is detected by osmoreceptors in the hypothalamus, which leads to a decrease in the secretion of ADH from the pituitary gland. A lower concentration of ADH in the blood makes the walls of the collecting ducts in the nephrons less permeable to water, resulting in less water reabsorption and the production of a large volume of dilute urine.
評分準則
1 mark for the correct answer: D. Any other option selected receives 0 marks.
題目 2 · 選擇題
1 分
A person drinks a large volume of water. Which of the following correctly describes the subsequent changes in the concentration of anti-diuretic hormone (ADH) in the blood and the permeability of the collecting ducts in the kidneys?
A.ADH concentration increases, and the permeability of the collecting ducts increases
B.ADH concentration increases, and the permeability of the collecting ducts decreases
C.ADH concentration decreases, and the permeability of the collecting ducts increases
D.ADH concentration decreases, and the permeability of the collecting ducts decreases
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解題
When a person drinks a large volume of water, the blood plasma becomes more dilute. This change is detected by osmoreceptors in the hypothalamus, which leads to a decrease in the secretion of ADH from the pituitary gland. A lower concentration of ADH in the blood makes the walls of the collecting ducts in the nephrons less permeable to water, resulting in less water reabsorption and the production of a large volume of dilute urine.
評分準則
1 mark for the correct answer: D. Any other option selected receives 0 marks.
題目 3 · Sentence Completion
4 分
Complete the following passage about the homeostatic control of water balance by writing the most appropriate word or words in each blank: When the concentration of water in the blood is low, this change is detected by receptor cells in the (i) _________________. This stimulates the pituitary gland to release more (ii) _________________ hormone into the blood. This hormone travels to the kidneys, where it increases the permeability of the collecting ducts. As a result, more water is reabsorbed back into the blood, and a small volume of (iii) _________________ urine is produced. This corrective mechanism is an example of (iv) _________________ feedback.
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解題
(i) The hypothalamus contains osmoreceptors that detect low water potential (or water concentration) in the blood. (ii) The hypothalamus signals the posterior pituitary gland to secrete antidiuretic hormone (ADH). (iii) ADH increases the permeability of the collecting ducts to water, allowing more water to be reabsorbed into the blood via osmosis. This leaves a smaller, more concentrated volume of urine in the collecting duct. (iv) Because this mechanism reverses the original change (low blood water concentration) to restore normal levels, it is an example of negative feedback.
評分準則
1 mark for each correct word: (i) hypothalamus (accept osmoregulatory centre); (ii) antidiuretic / ADH; (iii) concentrated (accept hypertonic); (iv) negative.
題目 4 · Sentence Completion
4 分
Complete the following passage about the homeostatic control of water balance by writing the most appropriate word or words in each blank: When the concentration of water in the blood is low, this change is detected by receptor cells in the (i) _________________. This stimulates the pituitary gland to release more (ii) _________________ hormone into the blood. This hormone travels to the kidneys, where it increases the permeability of the collecting ducts. As a result, more water is reabsorbed back into the blood, and a small volume of (iii) _________________ urine is produced. This corrective mechanism is an example of (iv) _________________ feedback.
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解題
(i) The hypothalamus contains osmoreceptors that detect low water potential (or water concentration) in the blood. (ii) The hypothalamus signals the posterior pituitary gland to secrete antidiuretic hormone (ADH). (iii) ADH increases the permeability of the collecting ducts to water, allowing more water to be reabsorbed into the blood via osmosis. This leaves a smaller, more concentrated volume of urine in the collecting duct. (iv) Because this mechanism reverses the original change (low blood water concentration) to restore normal levels, it is an example of negative feedback.
評分準則
1 mark for each correct word: (i) hypothalamus (accept osmoregulatory centre); (ii) antidiuretic / ADH; (iii) concentrated (accept hypertonic); (iv) negative.
題目 5 · Calculation
2 分
A student measures their resting heart rate as \(75\text{ beats per minute}\). Their stroke volume is estimated to be \(68\text{ cm}^3\). Calculate the cardiac output of this student in \(dm^3\text{ per minute}\).
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解題
Cardiac output is calculated using the formula: \(\text{Cardiac output} = \text{stroke volume} \times \text{heart rate}\). First, multiply the stroke volume by the heart rate to find the value in \(\text{cm}^3\text{ min}^{-1}\): \(68 \times 75 = 5100\text{ cm}^3\text{ min}^{-1}\). Next, convert the volume from \(\text{cm}^3\) to \(\text{dm}^3\) by dividing by 1000: \(5100 / 1000 = 5.1\text{ dm}^3\text{ min}^{-1}\).
評分準則
1 mark for correct working showing multiplication and division by 1000: \((68 \times 75) / 1000\) or finding \(5100\text{ cm}^3\text{ min}^{-1}\). 1 mark for the correct final answer of \(5.1\text{ dm}^3\text{ min}^{-1}\) (allow 5.1 alone).
題目 6 · Calculation
2 分
During mild exercise, a person has a tidal volume of \(0.65\text{ dm}^3\) and a breathing rate of \(16\text{ breaths per minute}\). Calculate their minute ventilation volume in \(dm^3\text{ min}^{-1}\).
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解題
Minute ventilation volume is calculated using the formula: \(\text{Minute ventilation} = \text{tidal volume} \times \text{breathing rate}\). Substitute the given values into the formula: \(0.65\text{ dm}^3 \times 16\text{ breaths/min} = 10.4\text{ dm}^3\text{ min}^{-1}\).
評分準則
1 mark for correct working showing multiplication: \(0.65 \times 16\). 1 mark for correct final value of \(10.4\text{ dm}^3\text{ min}^{-1}\) (allow 10.4 alone).
題目 7 · Calculation
2 分
A student uses a simple calorimeter to determine the energy content of a food sample. The mass of water used in the calorimeter is \(20\text{ g}\). The temperature of the water increases from \(19.0\ ^\circ\text{C}\) to \(44.0\ ^\circ\text{C}\) when the food sample is burned. Calculate the energy released in Joules (J). The specific heat capacity of water is \(4.2\text{ J/g}^\circ\text{C}\).
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解題
First, determine the rise in temperature of the water: \(44.0\ ^\circ\text{C} - 19.0\ ^\circ\text{C} = 25.0\ ^\circ\text{C}\). Next, calculate the energy released using the calorimetry formula: \(\text{Energy (J)} = \text{mass of water (g)} \times \text{specific heat capacity} \times \text{temperature change}\). Substitute the values: \(20 \times 4.2 \times 25.0 = 2100\text{ J}\).
評分準則
1 mark for calculating the correct temperature difference of \(25\ ^\circ\text{C}\) or showing the correct substitution: \(20 \times 4.2 \times 25\). 1 mark for the correct final answer of \(2100\text{ J}\) (allow 2100 alone).
題目 8 · Diagram Drawing / Labeling
3 分
A student is asked to label a diagram of a kidney nephron. Three structures are indicated by the labels X, Y, and Z.
- X is the cup-like sac at the beginning of the tubular component of a nephron. - Y is the U-shaped portion of the tubule that conducts urine within each nephron. - Z is the tube that receives urine from multiple nephrons and drains into the renal pelvis.
Identify the anatomical names of structures X, Y, and Z.
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解題
X represents the Bowman's capsule (or glomerular capsule), which performs the first step in the filtration of blood from the glomerulus to form urine. Y represents the Loop of Henle, which creates a high salt concentration in the medulla of the kidney to allow water reabsorption. Z represents the collecting duct, which carries the urine to the renal pelvis and is the site where ADH acts to regulate water reabsorption.
評分準則
- 1 mark for identifying X as the Bowman's capsule (accept glomerular capsule). - 1 mark for identifying Y as the Loop of Henle. - 1 mark for identifying Z as the collecting duct.
題目 9 · Diagram Drawing / Labeling
3 分
A diagram of an alveolus surrounded by a blood capillary is used to show the pathway of gas exchange in the lungs. Three parts of the diagram are labeled A, B, and C.
- A is the thin layer of water/moisture on the inside surface of the alveolus. - B is the single layer of cells that forms the wall of the alveolus. - C is the single layer of cells that forms the wall of the capillary.
State the names of structures/layers A, B, and C.
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解題
A represents the moisture layer (or fluid lining/surfactant), which dissolves oxygen before it diffuses across the membranes. B represents the alveolar wall (or alveolar epithelium / squamous epithelium), which is one-cell thick to minimize diffusion distance. C represents the capillary wall (or capillary endothelium / endothelial cell layer), which is also one-cell thick to allow rapid diffusion of gases into the blood.
評分準則
- 1 mark for identifying A as the moisture layer / layer of moisture (accept fluid lining / water film / surfactant). - 1 mark for identifying B as the alveolar wall / alveolar epithelium (accept squamous epithelium). - 1 mark for identifying C as the capillary wall / capillary endothelium (accept endothelial layer).
題目 10 · Diagram Drawing / Labeling
3 分
The diagram below represents a synovial joint (such as the elbow joint).
Three parts are labeled D, E, and F. - D is a band of strong, fibrous connective tissue that connects bone to bone. - E is a smooth, slippery layer of tissue that covers the articulating ends of the bones. - F is the membrane that encloses the joint cavity and secretes a lubricating fluid.
Identify the structures D, E, and F.
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解題
D is a ligament, which connects bones to other bones and stabilizes the joint. E is articular cartilage, which reduces friction and absorbs shock between the moving bones. F is the synovial membrane, which secretes synovial fluid to lubricate the joint cavity and minimize friction.
評分準則
- 1 mark for identifying D as ligament. - 1 mark for identifying E as cartilage (accept articular cartilage). - 1 mark for identifying F as the synovial membrane.
題目 11 · Diagram Drawing / Labeling
3 分
A diagram shows a fetus developing inside the uterus. Three key structures are labeled J, K, and L.
- J is the membrane/sac that encloses the developing fetus. - K is the liquid surrounding the fetus that cushions it against physical impacts. - L is the temporary organ that connects the developing fetus to the uterine wall, allowing nutrient uptake, waste elimination, and gas exchange.
Name the structures or liquid labeled J, K, and L.
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解題
J represents the amniotic sac (or amnion), which develops to enclose the embryo/fetus. K represents the amniotic fluid, which serves as a cushion to protect the developing fetus from mechanical shock, maintains a constant temperature, and allows room for growth. L represents the placenta, which facilitates the exchange of oxygen, glucose, antibodies, and waste products between the maternal and fetal bloodstreams.
評分準則
- 1 mark for identifying J as the amniotic sac (accept amnion). - 1 mark for identifying K as the amniotic fluid. - 1 mark for identifying L as the placenta.
題目 12 · Short Answer
2 分
A student burns a 0.50 g sample of dried bread under a boiling tube containing 20.0 g of water. The temperature of the water rises from 20.0 °C to 35.0 °C. Calculate the energy released per gram of this food sample in joules per gram (J/g). (Specific heat capacity of water = 4.2 J/g/°C)
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解題
Step 1: Calculate the total heat energy absorbed by the water using the formula \(Q = m \times c \times \Delta T\). Here, \(m = 20.0\text{ g}\), \(c = 4.2\text{ J/g/°C}\), and \(\Delta T = 35.0 - 20.0 = 15.0\text{ °C}\). So, \(Q = 20.0 \times 4.2 \times 15.0 = 1260\text{ J}\). Step 2: Calculate the energy per gram by dividing the total energy by the mass of the food sample: \(\text{Energy per gram} = 1260\text{ J} / 0.50\text{ g} = 2520\text{ J/g}\).
評分準則
1 mark for calculating the total energy correctly as 1260 J (or showing the correct working: \(20 \times 4.2 \times 15\)). 1 mark for dividing by the mass of food to get the correct final value of 2520 J/g.
題目 13 · Short Answer
2 分
Name the cell organelle responsible for aerobic respiration, and identify the specific folded inner membrane structures within this organelle that increase its surface area.
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解題
The mitochondrion is the site of aerobic respiration where ATP is produced. The inner membrane of a mitochondrion is highly folded into structures called cristae, which increase the surface area available for the chemical reactions of respiration.
評分準則
1 mark for naming mitochondrion / mitochondria. 1 mark for naming cristae (accept crista).
題目 14 · Short Answer
2 分
Distinguish between a pathogen and a vector by defining each term using malaria as an illustrative example.
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解題
A pathogen is defined as an organism that causes disease, such as the protoctist Plasmodium which causes malaria. A vector is an organism that carries and transmits a pathogen into another living organism without causing the disease directly itself, such as the female Anopheles mosquito.
評分準則
1 mark for defining pathogen with reference to Plasmodium as the causative agent. 1 mark for defining vector with reference to the Anopheles mosquito as the transmitter.
題目 15 · Short Answer
2 分
Describe what happens to a red blood cell when it is placed into a hypertonic (highly concentrated) salt solution, and state the term used to describe its final appearance.
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解題
When placed in a hypertonic solution, the water potential outside the red blood cell is lower than inside. Water leaves the cell by osmosis down a water potential gradient across the selectively permeable membrane. This causes the cell to shrink and shrivel, a process called crenation.
評分準則
1 mark for explaining that water leaves the cell by osmosis / down a water potential gradient. 1 mark for stating that the cell shrinks/shrivels or undergoes crenation.
題目 16 · Short Answer
2 分
Identify the endocrine gland and the specific cells within it that secrete the hormone glucagon, and name the main target organ of this hormone.
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解題
Glucagon is synthesized and secreted by the alpha cells located in the Islets of Langerhans of the pancreas. Its primary target organ is the liver, where it stimulates the breakdown of glycogen into glucose (glycogenolysis).
評分準則
1 mark for pancreas (accept Islets of Langerhans or alpha cells). 1 mark for liver.
題目 17 · Short Answer
2 分
State the name of the chemical reagent used to test for the presence of a reducing sugar, and describe the positive colour change observed when a high concentration of reducing sugar is present.
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解題
To test for reducing sugars, Benedict's reagent is added to the food sample and heated in a water bath. If a high concentration of reducing sugar is present, the colour changes from the original blue to a brick-red precipitate.
評分準則
1 mark for identifying Benedict's reagent (or Benedict's solution). 1 mark for stating the final colour is brick-red (accept red / orange-red; reject yellow or green as indicative of a high concentration).
題目 18 · Short Answer
2 分
Name the pigmented, muscular structure in the human eye that regulates the amount of light entering the pupil, and identify the two groups of antagonistic muscles it contains.
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解題
The iris is the colored muscular diaphragm that controls the size of the pupil. It consists of two sets of smooth muscles that work antagonistically to control pupil diameter: circular muscles and radial muscles.
評分準則
1 mark for naming the iris. 1 mark for identifying both circular and radial muscles.
題目 19 · Short Answer
2 分
State the sex chromosome configuration of a human male, and determine the probability of a couple having a male child.
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解題
A human male has the sex chromosome configuration XY. During fertilization, there is a 50% (or 0.5 / 1 in 2) chance that a sperm carrying a Y chromosome fertilizes the egg (which always carries an X chromosome), resulting in a male offspring.
評分準則
1 mark for stating the male sex chromosomes are XY. 1 mark for stating the probability is 0.5 / 50% / 1 in 2.
題目 20 · Short Answer
2 分
State the gland that secretes antidiuretic hormone (ADH) and describe its effect on the permeability of the collecting duct of a nephron.
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解題
Antidiuretic hormone (ADH) is released into the bloodstream by the pituitary gland. It travels to the kidneys where it acts on the collecting ducts of the nephrons, increasing their permeability to water. This allows more water to be reabsorbed back into the blood plasma by osmosis, helping to conserve water in the body.
評分準則
1. Pituitary gland (1 mark) 2. Increases permeability of the collecting duct to water / allows more water to be reabsorbed (1 mark)
題目 21 · Short Answer
2 分
Describe how a student can test a liquid food sample to show that it contains a non-reducing sugar, such as sucrose.
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解題
To test for a non-reducing sugar like sucrose, first add dilute hydrochloric acid to the liquid sample and heat in a water bath to hydrolyse the disaccharide into reducing sugars. Next, neutralise the acid by adding an alkali such as sodium hydrogencarbonate. Finally, add Benedict's reagent and heat the mixture again in a water bath. A colour change from blue to green, yellow, orange, or brick-red precipitate confirms the presence of non-reducing sugars.
評分準則
1. Heat sample with dilute hydrochloric acid AND neutralise with an alkali / sodium hydrogencarbonate (1 mark) 2. Add Benedict's reagent and heat, resulting in a colour change from blue to green/yellow/orange/brick-red (1 mark)
題目 22 · Extended Answer
4 分
Describe how the structure of ciliated epithelial cells in the respiratory tract is adapted to their function.
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解題
Ciliated epithelial cells have hair-like projections called cilia on their free surface. These cilia beat rhythmically and in a coordinated manner to move mucus, which has trapped dust and pathogens, upwards and out of the respiratory tract. To support this active movement, these cells contain a high density of mitochondria that generate ATP through aerobic respiration. Additionally, they work alongside mucus-secreting goblet cells to maintain a clean airway.
評分準則
1. Mention of cilia/hair-like structures sweeping or moving mucus (1 mark). 2. Pathogens/dust trapped in mucus are moved away from lungs/towards throat (1 mark). 3. High density of mitochondria present (1 mark). 4. To provide ATP/energy required for cilia to beat (1 mark).
題目 23 · Extended Answer
4 分
Explain how active transport differs from facilitated diffusion, referencing the movement of glucose in the human nephron.
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解題
Active transport requires metabolic energy in the form of ATP to transport substances against their concentration gradient (from a low concentration to a high concentration) across a cell membrane. Facilitated diffusion is a passive process that does not require energy, moving substances down their concentration gradient using specific carrier or channel proteins. In the proximal convoluted tubule of the nephron, glucose must be completely reabsorbed into the blood. This requires active transport proteins to pump glucose from the low concentration in the filtrate to the higher concentration in the tubule cells.
評分準則
1. Active transport moves substances against a concentration gradient, while facilitated diffusion moves them down a gradient (1 mark). 2. Active transport requires ATP/energy, whereas facilitated diffusion is passive/no energy required (1 mark). 3. Identify that glucose is selectively reabsorbed in the proximal convoluted tubule of the nephron (1 mark). 4. Explain that active transport ensures all glucose is retrieved even when its concentration in the filtrate is very low (1 mark).
題目 24 · Extended Answer
4 分
Explain how the structural differences between an artery and a vein relate to the pressure of the blood they transport.
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解題
Arteries transport blood under high pressure away from the heart. Consequently, they have thick walls containing abundant elastic fibers and smooth muscle. The elastic fibers stretch and recoil to withstand and maintain high blood pressure, while the muscular layer helps regulate flow. In contrast, veins carry blood under much lower pressure back to the heart, so they have thinner walls and a wider lumen to reduce resistance. Because the pressure is low, veins also contain one-way valves to prevent the backflow of blood.
評分準則
1. Arteries have thick muscular/elastic walls to withstand and maintain high pressure (1 mark). 2. Elastic fibers in arteries stretch and recoil to smooth out high pressure blood flow (1 mark). 3. Veins have thinner walls and a larger lumen because they transport low-pressure blood (1 mark). 4. Veins possess valves to prevent backflow under low pressure (1 mark).
題目 25 · Extended Answer
4 分
Describe the sequence of physical changes that occur in the human thorax during inhalation.
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解題
During inhalation, the external intercostal muscles contract, pulling the ribcage upwards and outwards. Simultaneously, the diaphragm muscles contract, causing the diaphragm to flatten and move downwards. These two actions significantly increase the volume of the thoracic cavity. As a result of this volume increase, the air pressure inside the lungs drops below atmospheric pressure, forcing air to rush in through the trachea and bronchi to equalize the pressure.
評分準則
1. External intercostal muscles contract and pull the ribcage up and out (1 mark). 2. Diaphragm contracts and flattens/moves down (1 mark). 3. Volume of the thoracic cavity increases (1 mark). 4. Pressure inside the lungs decreases below atmospheric pressure, drawing air in (1 mark).
題目 26 · Extended Answer
4 分
Explain how the human body coordinates physiological responses to lower internal body temperature when it exceeds normal levels.
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解題
When internal body temperature rises, the thermoregulatory center in the hypothalamus detects the change and coordinates a response. First, vasodilation occurs, where arterioles supplying skin capillaries dilate, allowing more warm blood to flow near the skin surface to lose heat by radiation. Second, sweat glands are stimulated to secrete sweat, which evaporates from the skin, removing thermal energy. Third, hair erector muscles relax, causing hairs to lie flat and preventing a layer of warm insulating air from being trapped next to the skin.
評分準則
1. Hypothalamus/thermoreceptors detect temperature rise (1 mark). 2. Vasodilation occurs where skin arterioles dilate to increase blood flow near surface, increasing heat loss by radiation (1 mark). 3. Sweat glands secrete sweat, leading to cooling via evaporation (1 mark). 4. Hair erector muscles relax so hairs lie flat, reducing insulation (1 mark).
題目 27 · Extended Answer
4 分
Describe how an electrical impulse is transmitted across a synapse from a presynaptic neuron to a postsynaptic neuron.
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解題
When an electrical impulse reaches the presynaptic knob, it triggers synaptic vesicles to fuse with the presynaptic membrane, releasing neurotransmitter chemical substances into the synaptic cleft via exocytosis. These neurotransmitters diffuse across the gap down their concentration gradient. Upon reaching the postsynaptic membrane, they bind specifically to complementary receptor molecules. This binding causes ion channels to open, generating a new electrical impulse in the postsynaptic neuron.
評分準則
1. Electrical impulse reaches presynaptic terminal, causing vesicles to fuse with membrane (1 mark). 2. Neurotransmitters are released into the synaptic cleft (1 mark). 3. Neurotransmitters diffuse across the cleft (1 mark). 4. Neurotransmitters bind to specific receptors on the postsynaptic membrane, triggering a new impulse (1 mark).
題目 28 · Extended Answer
4 分
Explain why a person who has previously been infected with a pathogen does not usually fall ill upon a second exposure to the same pathogen.
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解題
During the primary immune response, specific lymphocytes are activated and some differentiate into memory cells, which persist in the blood and lymphatic system. Upon a second exposure to the same pathogen, these memory cells immediately recognize the foreign antigens. They undergo rapid cell division (clonal expansion) to produce a large population of active plasma cells. These plasma cells produce and secrete the specific antibodies much faster and in significantly higher quantities, neutralizing the pathogen before it can multiply sufficiently to cause symptoms.
評分準則
1. Memory cells (B/T lymphocytes) remain in the body from the primary infection (1 mark). 2. On re-exposure, memory cells rapidly recognize the specific antigen (1 mark). 3. Rapid division occurs to produce many plasma cells (1 mark). 4. Specific antibodies are produced much faster and in greater quantities, destroying the pathogen before symptoms occur (1 mark).
題目 29 · Extended Answer
4 分
Explain how the structure of the placenta is adapted to facilitate efficient exchange of substances between maternal and fetal blood.
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解題
The placenta has several key structural adaptations to maximize transport. First, it contains numerous finger-like projections called chorionic villi, which provide an extremely large surface area for diffusion. Second, the barrier between maternal and fetal blood is incredibly thin, consisting of just a single layer of epithelial cells, which ensures a short diffusion pathway. Third, a highly developed network of fetal capillaries and a continuous supply of maternal blood maintain a steep concentration gradient, ensuring the rapid and efficient movement of oxygen, glucose, and waste products like carbon dioxide.
評分準則
1. Chorionic villi provide a large surface area for efficient diffusion (1 mark). 2. Very thin barrier (one cell thick) ensures a short diffusion distance (1 mark). 3. Rich blood supply in fetal capillaries maintains a steep concentration gradient (1 mark). 4. Close proximity of maternal blood pools allows exchange of oxygen/nutrients and waste products (1 mark).
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