An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V1) Cambridge International A Level Biology (0610) paper. Not affiliated with or reproduced from Cambridge.
卷二: 選擇題 (Extended)
There are forty questions on this paper. Answer all questions. For each question there are four possible answers. Choose the one you consider correct.
40 題目 · 40 分
題目 1 · 選擇題
1 分
Four identical cylinders of potato, each with an initial mass of 5.0 g, were placed into sucrose solutions of different concentrations (0.1, 0.3, 0.5 and 0.7 mol/dm³). After two hours, they were reweighed. The final masses recorded were 5.4 g, 5.0 g, 4.6 g and 4.2 g respectively. Which concentration is closest to the sucrose concentration inside the potato cell vacuoles?
A.0.1 mol/dm³
B.0.3 mol/dm³
C.0.5 mol/dm³
D.0.7 mol/dm³
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解題
At 0.3 mol/dm³, the final mass of the potato cylinder remains unchanged at 5.0 g. This indicates that there is no net movement of water into or out of the potato cells by osmosis because the water potential of the external sucrose solution is equal to the water potential inside the potato vacuoles.
評分準則
Award 1 mark for the correct option (B). No marks are awarded for incorrect options.
題目 2 · 選擇題
1 分
A photomicrograph shows a plant cell. The measured length of the cell in the photograph is 48 mm. If the magnification of the image is \(\times 800\), what is the actual length of the plant cell?
A.0.06 \(\mu\text{m}\)
B.0.6 \(\mu\text{m}\)
C.60 \(\mu\text{m}\)
D.600 \(\mu\text{m}\)
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解題
Using the magnification formula: \(\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}\). First, convert the image size from millimetres to micrometres: \(48\text{ mm} = 48,000\ \mu\text{m}\). Then, divide by the magnification: \(\frac{48,000\ \mu\text{m}}{800} = 60\ \mu\text{m}\).
評分準則
Award 1 mark for the correct option (C). No marks are awarded for incorrect options.
題目 3 · 選擇題
1 分
A student sets up a potometer to investigate the effect of environmental factors on the rate of transpiration in a leafy shoot. Which set of conditions will result in the fastest rate of movement of the air bubble in the capillary tube?
A.High temperature, high humidity, high wind speed
B.High temperature, low humidity, high wind speed
C.Low temperature, high humidity, low wind speed
D.Low temperature, low humidity, high wind speed
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解題
Transpiration occurs fastest under conditions of high temperature (which increases the kinetic energy of water molecules), low humidity (which maintains a steep water potential gradient between the leaf interior and the air), and high wind speed (which sweeps away water vapour from the leaf surface, maintaining the gradient). A faster transpiration rate results in more rapid water uptake and faster movement of the air bubble.
評分準則
Award 1 mark for the correct option (B). No marks are awarded for incorrect options.
題目 4 · 選擇題
1 分
In a species of flowering plant, flower colour is controlled by a single gene with two codominant alleles, \(C^R\) (red flowers) and \(C^W\) (white flowers). Heterozygous plants (\(C^R C^W\)) have pink flowers. If two pink-flowered plants are crossed, what is the expected phenotypic ratio of their offspring?
A.3 red : 1 white
B.1 red : 2 pink : 1 white
C.1 red : 1 white
D.All pink
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解題
The cross is \(C^R C^W \times C^R C^W\). The offspring genotypes will be 25% \(C^R C^R\) (red), 50% \(C^R C^W\) (pink), and 25% \(C^W C^W\) (white). This gives an expected phenotypic ratio of 1 red : 2 pink : 1 white.
評分準則
Award 1 mark for the correct option (B). No marks are awarded for incorrect options.
題目 5 · 選擇題
1 分
Which of the following is an example of passive immunity?
A.Production of antibodies by lymphocytes after a pathogen enters the body
B.Injection of a weakened pathogen during vaccination
C.Transfer of maternal antibodies across the placenta to a fetus
D.Production of memory cells after recovering from an infection
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解題
Passive immunity involves receiving antibodies from another individual, rather than producing them yourself. The transfer of antibodies from mother to fetus across the placenta provides temporary, passive protection. Options A, B, and D describe active immunity, where the body's own immune system produces antibodies and memory cells.
評分準則
Award 1 mark for the correct option (C). No marks are awarded for incorrect options.
題目 6 · 選擇題
1 分
Pectinase is an enzyme widely used in the industrial production of fruit juices. What is the primary biological reason for adding pectinase to crushed fruit?
A.To break down cell walls to increase juice yield and clarity
B.To ferment fruit sugars into ethanol and carbon dioxide
C.To denature bacterial enzymes and preserve the juice
D.To convert starch molecules into glucose to sweeten the product
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解題
Pectin is a structural polysaccharide found in plant cell walls. Adding pectinase breaks down this pectin, weakening the cell walls and allowing more juice to be extracted easily, while also helping to clarify the juice by breaking down suspended particles.
評分準則
Award 1 mark for the correct option (A). No marks are awarded for incorrect options.
題目 7 · 選擇題
1 分
Which hormone is responsible for maintaining the uterine lining during early pregnancy, and which structure primarily secretes it before the placenta is fully developed?
A.Follicle-stimulating hormone (FSH), secreted by the pituitary gland
B.Estrogen, secreted by the ovaries
C.Luteinising hormone (LH), secreted by the corpus luteum
D.Progesterone, secreted by the corpus luteum
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解題
Progesterone is the hormone responsible for maintaining the thick, vascular lining of the uterus during early pregnancy. After ovulation, the remains of the follicle develop into the corpus luteum, which secretes progesterone to maintain the pregnancy until the placenta takes over this function.
評分準則
Award 1 mark for the correct option (D). No marks are awarded for incorrect options.
題目 8 · 選擇題
1 分
A student accidentally touches a very hot pan, causing a reflex withdrawal of the hand. Which pathway represents the correct flow of nerve impulses during this reflex action?
The reflex arc begins when a receptor detects a stimulus. The nerve impulse is carried along a sensory neurone to the central nervous system (spinal cord). Within the spinal cord, the impulse is transmitted across a synapse to a relay neurone, and then across another synapse to a motor neurone. The motor neurone carries the impulse to the effector (muscle), which carries out the response.
評分準則
Award 1 mark for the correct option (B). No marks are awarded for incorrect options.
題目 9 · 選擇題
1 分
A sample of double-stranded DNA was analyzed and found to contain 34% cytosine. What is the percentage of adenine in this DNA sample?
A.16%
B.34%
C.68%
D.32%
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解題
In double-stranded DNA, the bases pair specifically: cytosine (C) pairs with guanine (G), and adenine (A) pairs with thymine (T). Since cytosine is 34%, guanine must also be 34%. Together, C + G = 68%. This leaves 32% of the total DNA for adenine and thymine combined (100% - 68% = 32%). Since the amount of adenine is equal to the amount of thymine, the percentage of adenine is 32% / 2 = 16%.
評分準則
1 mark for selecting the correct percentage of adenine (A).
題目 10 · 選擇題
1 分
Four similar cylinders of potato tissue, each of initial length 50 mm, were placed in different concentrations of sucrose solution. After 24 hours, their final lengths were measured: Solution P = 46 mm, Solution Q = 52 mm, Solution R = 49 mm, Solution S = 55 mm. Which sucrose solution has a water potential closest to that of the potato cells?
A.Solution P
B.Solution Q
C.Solution R
D.Solution S
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解題
The solution with a water potential closest to that of the potato cells will cause the least net movement of water by osmosis. This results in the smallest change in length. The changes in length are: Solution P = -4 mm; Solution Q = +2 mm; Solution R = -1 mm; Solution S = +5 mm. The smallest overall change is -1 mm with Solution R, meaning its water potential is closest to that of the cells.
評分準則
1 mark for identifying the solution with the smallest change in length (C).
題目 11 · 選擇題
1 分
Which row correctly describes passive immunity?
A.Antibodies are made by the body itself; Memory cells are produced; Protection is long-term.
B.Antibodies are acquired from another individual; Memory cells are not produced; Protection is short-term.
C.Antibodies are made by the body itself; Memory cells are not produced; Protection is short-term.
D.Antibodies are acquired from another individual; Memory cells are produced; Protection is long-term.
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解題
Passive immunity involves obtaining antibodies from another individual (e.g., across the placenta, via breast milk, or through an injection). Because the host's own immune system is not activated, no memory cells are produced, and the protection is temporary and short-term.
評分準則
1 mark for the correct combination of attributes representing passive immunity (B).
題目 12 · 選擇題
1 分
A leafy shoot is fitted into a potometer. The rate of movement of the air bubble is measured under different environmental conditions. Which set of conditions will produce the fastest rate of bubble movement?
A.High temperature, high humidity, high wind speed
B.High temperature, low humidity, high wind speed
C.Low temperature, low humidity, low wind speed
D.High temperature, low humidity, low wind speed
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解題
The rate of bubble movement represents the rate of transpiration. Transpiration is fastest when the water potential gradient between the inside of the leaf and the external air is steepest. This is achieved by: 1) High temperature, which increases the kinetic energy of water molecules and rate of evaporation; 2) Low humidity, which maintains a dry atmosphere outside the leaf; and 3) High wind speed, which sweeps away accumulated water vapor from around the stomata.
評分準則
1 mark for the correct set of environmental factors (B).
題目 13 · 選擇題
1 分
In a breed of cats, black fur is dominant to chocolate fur. A homozygous black male is mated with a chocolate female. Two offspring from this cross are mated with each other. What is the expected phenotypic ratio of black to chocolate fur in their offspring?
A.All black
B.1 black : 1 chocolate
C.3 black : 1 chocolate
D.All chocolate
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解題
Let B represent the dominant allele for black fur, and b represent the recessive allele for chocolate fur. The homozygous black male (BB) and chocolate female (bb) produce offspring that are all heterozygous (Bb) and have black fur. When two of these heterozygous F1 offspring are crossed (Bb x Bb), the expected F2 genotypic ratio is 1 BB : 2 Bb : 1 bb. Since B is dominant, BB and Bb are black, and bb is chocolate. This gives a phenotypic ratio of 3 black : 1 chocolate.
評分準則
1 mark for the correct F2 phenotypic ratio (C).
題目 14 · 選擇題
1 分
An industrial fermenter is used to grow penicillin-producing fungi. Why is it necessary to bubble sterile air into the fermenter during this process?
A.To provide carbon dioxide for photosynthesis and keep the temperature high.
B.To provide oxygen for aerobic respiration and to help keep the culture mixed.
C.To prevent the mixture from becoming too acidic by removing oxygen.
D.To cool the fermenter down and prevent any anaerobic respiration.
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解題
Penicillium is a fungus that requires oxygen for aerobic respiration to grow and produce penicillin. Bubbling sterile air provides this essential oxygen. Additionally, the physical movement of the air bubbles rising through the liquid medium helps to stir and mix the culture, ensuring nutrients and temperature are evenly distributed.
評分準則
1 mark for identifying the correct functions of sterile air in a fermenter (B).
題目 15 · 選擇題
1 分
Which method is most effective at preventing the transmission of the Human Immunodeficiency Virus (HIV) during sexual intercourse?
A.Taking a chemical contraceptive pill daily
B.Using a latex condom
C.Using the rhythm method to monitor body temperature
D.Using a spermicidal gel
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解題
HIV is transmitted through direct contact with infected bodily fluids, such as semen and vaginal fluids. A latex condom acts as a mechanical barrier that prevents these fluids from being exchanged during sexual intercourse. Other methods, such as chemical or natural methods of contraception, do not prevent fluid exchange and offer no protection against HIV transmission.
評分準則
1 mark for selecting the physical barrier method (B).
題目 16 · 選擇題
1 分
A person accidentally touches a hot stove and quickly pulls their hand away. Which sequence shows the correct pathway of nerve impulses in this reflex action?
In a reflex arc, a stimulus is detected by a receptor. The receptor initiates a nerve impulse that travels along a sensory neurone to the central nervous system (spinal cord). Inside the spinal cord, the impulse is passed across a synapse to a relay neurone, and then across another synapse to a motor neurone. The motor neurone carries the impulse to the effector (a muscle), which carries out the response.
評分準則
1 mark for the correct chronological sequence of components in a reflex arc (C).
題目 17 · 選擇題
1 分
A student measures the length of a plant cell on a micrograph. The measured length of the cell on the micrograph is 48 mm. The actual length of the cell is 80 \mu m. What is the magnification of the micrograph?
A.\times 0.6
B.\times 600
C.\times 60
D.\times 6000
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解題
To calculate the magnification, use the formula: \text{Magnification} = \frac{\text{Image size (I)}}{\text{Actual size (A)}} First, convert both measurements to the same units. Converting 48 mm to micrometres (\mu m): 48 mm = 48 \times 1000 = 48,000 \mu m. Now, divide the image size by the actual size: \text{Magnification} = \frac{48,000 \mu m}{80 \mu m} = 600. Therefore, the magnification is \times 600.
評分準則
1 mark for the correct option (B). Award 1 mark if the candidate correctly calculates the magnification using the converted values.
題目 18 · 選擇題
1 分
Plant cells are placed in a concentrated sucrose solution that has a lower water potential than the cell cytoplasm. Which row correctly describes what happens to the vacuole volume and the cell membrane?
A.Vacuole volume decreases; cell membrane pulls away from the cell wall
B.Vacuole volume decreases; cell membrane remains pressed against the cell wall
C.Vacuole volume increases; cell membrane pulls away from the cell wall
D.Vacuole volume increases; cell membrane remains pressed against the cell wall
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解題
The concentrated sucrose solution is hypertonic to the cytoplasm, meaning it has a lower water potential. Water leaves the plant cell vacuole and cytoplasm by osmosis down the water potential gradient, through the partially permeable cell membrane. This causes the vacuole volume to decrease and the cytoplasm/cell membrane to shrink and pull away from the cell wall, a process called plasmolysis.
評分準則
1 mark for the correct option (A). Award 1 mark for identifying that water loss causes the vacuole to shrink and the membrane to detach from the wall.
題目 19 · 選擇題
1 分
Which features are present in mature, functioning xylem vessels?
1. Cytoplasm 2. Lignified cell walls 3. Companion cells 4. End walls completely broken down
A.1 and 2
B.1 and 3
C.2 and 4
D.3 and 4
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解題
Mature xylem vessels are dead, hollow tubes that have no cytoplasm (1 is incorrect) and no end walls (4 is correct), allowing an uninterrupted column of water to flow. Their cell walls are reinforced with lignin (2 is correct) to withstand the negative pressure of transpiration. Companion cells (3) are associated with phloem sieve tubes, not xylem.
評分準則
1 mark for the correct option (C). Award 1 mark for identifying that xylem lacks cytoplasm/end walls and has lignified walls.
題目 20 · 選擇題
1 分
In a breed of cats, black fur is caused by a dominant allele (B) and white fur is caused by a recessive allele (b). Two black cats are mated and produce a litter of six kittens: four are black and two are white. What are the genotypes of the parents?
A.BB and BB
B.Bb and Bb
C.BB and Bb
D.Bb and bb
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解題
Since some kittens have white fur, they must have the homozygous recessive genotype (bb). Each white kitten must inherit one recessive allele (b) from each parent. Since both parents have black fur, they must also carry at least one dominant allele (B). Therefore, both parents must be heterozygous (Bb) to be black themselves and yet be able to produce white offspring.
評分準則
1 mark for the correct option (B). Award 1 mark for recognizing that producing white offspring requires both parents to carry the recessive allele.
題目 21 · 選擇題
1 分
Which statement describes a characteristic feature of passive immunity?
A.It is always long-term because memory cells are produced.
B.It is acquired only by artificial means, such as vaccination.
C.It involves the production of antibodies by the person's own lymphocytes.
D.It provides immediate protection but does not result in immunological memory.
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解題
Passive immunity is the temporary defense against a pathogen by antibody transfer from another source. Because the individual's own lymphocytes do not produce these antibodies, no memory cells are created, so there is no immunological memory, although protection is immediate. Passive immunity can also be natural, such as antibodies transferring across the placenta or through breast milk.
評分準則
1 mark for the correct option (D). Award 1 mark for identifying that passive immunity lacks memory cells but offers immediate protection.
題目 22 · 選擇題
1 分
In the industrial production of penicillin in a fermenter, what is the main purpose of bubbling sterile air through the nutrient broth?
A.To supply carbon dioxide for Penicillium to perform photosynthesis.
B.To provide oxygen for aerobic respiration of Penicillium and to keep the contents mixed.
C.To maintain a high temperature to speed up penicillin synthesis.
D.To prevent the pH of the mixture from falling below neutral.
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解題
Penicillium is a fungus that requires oxygen for aerobic respiration to grow and produce penicillin. Bubbling sterile air through the mixture supplies this oxygen. Furthermore, the rising bubbles help to mix the nutrient broth, ensuring that the fungus, nutrients, and temperature are uniformly distributed throughout the fermenter.
評分準則
1 mark for the correct option (B). Award 1 mark for stating that oxygen is needed for respiration and the air helps mix the contents.
題目 23 · 選擇題
1 分
Which cell type is targeted and destroyed by the Human Immunodeficiency Virus (HIV), leading to a reduction in the body's ability to produce antibodies?
A.Phagocytes
B.Lymphocytes
C.Platelets
D.Red blood cells
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解題
HIV infects and destroys lymphocytes (specifically helper T-cells). Since lymphocytes are responsible for antibody production and coordinating the immune response, their destruction severely weakens the immune system, eventually leading to AIDS.
評分準則
1 mark for the correct option (B). Award 1 mark for recognizing lymphocytes as the target of HIV.
題目 24 · 選擇題
1 分
The sequence shows the pathway of an electrical impulse along a reflex arc:
receptor \rightarrow sensory neurone \rightarrow synapse \rightarrow X \rightarrow synapse \rightarrow motor neurone \rightarrow effector
What is structure X?
A.Relay neurone
B.Sensory receptor
C.Brain coordinator
D.Effector organ
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解題
A reflex arc typically involves three types of neurone. The sensory neurone transmits the impulse from the receptor to the central nervous system (CNS), where it passes across a synapse to a relay neurone (structure X). The impulse then crosses another synapse to a motor neurone, which carries the signal to the effector.
評分準則
1 mark for the correct option (A). Award 1 mark for identifying structure X as the relay neurone linking sensory and motor neurones.
題目 25 · multiple_choice
1 分
A sample of red blood cells and a sample of palisade mesophyll cells are both placed in a container of distilled water. Which row correctly describes the state of the cells after 30 minutes?
Distilled water has a higher water potential than the cytoplasm of both cell types. Water enters both cells by osmosis down a water potential gradient. Red blood cells do not have a cell wall, so they swell and burst (lyse). Palisade mesophyll cells have a strong cellulose cell wall, which prevents them from bursting, making them turgid instead.
評分準則
1 mark for selecting correct option A. Correctly identifies that red blood cells burst and plant cells become turgid in pure water.
題目 26 · multiple_choice
1 分
A person accidentally touches a hot object with their hand. The reflex action that follows involves several structures. In what order does an electrical impulse travel through these structures?
The pathway of a reflex arc starts when a receptor (4) detects a stimulus. An electrical impulse travels along a sensory neurone (2) to the spinal cord (CNS). In the spinal cord, the impulse is transmitted across a synapse to a relay neurone (3), and then across another synapse to a motor neurone (1). The motor neurone carries the impulse to an effector (5) to produce a response. Thus, the correct sequence is 4 → 2 → 3 → 1 → 5.
評分準則
1 mark for selecting correct option A. Correctly identifies the complete pathway of impulses from receptor to effector through sensory, relay, and motor neurones.
題目 27 · multiple_choice
1 分
A mother has blood group A and has the genotype \(I^A I^O\). A father has blood group B and has the genotype \(I^B I^O\). What is the probability that their first child will have blood group O?
A.0%
B.25%
C.50%
D.75%
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解題
The mother can pass on either the \(I^A\) or \(I^O\) allele, and the father can pass on either the \(I^B\) or \(I^O\) allele. The possible genotypes for their offspring are: \(I^A I^B\) (blood group AB), \(I^A I^O\) (blood group A), \(I^B I^O\) (blood group B), and \(I^O I^O\) (blood group O). Out of these four possible outcomes, only \(I^O I^O\) results in blood group O. Thus, the probability is 1 out of 4, or 25%.
評分準則
1 mark for selecting correct option B. Correctly determines offspring genotypes using a Punnett square and calculates the probability of group O.
題目 28 · multiple_choice
1 分
Which row correctly identifies an example of active immunity and an example of passive immunity?
A.Active immunity: Production of antibodies after receiving a vaccine; Passive immunity: Transfer of antibodies across the placenta to a fetus
B.Active immunity: Absorption of antibodies from breast milk; Passive immunity: Production of memory cells after an infection
C.Active immunity: Injection of pre-made antibodies; Passive immunity: Vaccination against tuberculosis
D.Active immunity: Transfer of antibodies across the placenta; Passive immunity: Production of antibodies after an infection
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解題
Active immunity is defence against a pathogen by antibody production in the body, which is stimulated by a vaccine. Passive immunity is the short-term defence against a pathogen by antibodies acquired from another individual, such as the transfer of maternal antibodies across the placenta to a developing fetus.
評分準則
1 mark for selecting correct option A. Correctly distinguishes between active immunity (stimulated antibody production) and passive immunity (acquired antibodies).
題目 29 · multiple_choice
1 分
The rate of transpiration from a plant was measured under four different sets of environmental conditions. Under which set of conditions would the rate of transpiration be the highest?
A.Temperature: high, Humidity: high, Wind speed: high
B.Temperature: high, Humidity: low, Wind speed: high
Transpiration increases with higher temperatures because water molecules have more kinetic energy to evaporate. It increases with lower humidity because this maintains a steep water potential gradient between the air spaces inside the leaf and the external atmosphere. High wind speed sweeps away water vapor from the leaf surface, keeping the external air dry and maintaining this gradient. Therefore, high temperature, low humidity, and high wind speed result in the highest rate of transpiration.
評分準則
1 mark for selecting correct option B. Correctly identifies how temperature, humidity, and wind speed affect the transpiration rate.
題目 30 · multiple_choice
1 分
Yeast is used in the manufacture of bread. Which products of anaerobic respiration in yeast are essential for making bread rise, and what happens to the other major product?
A.Product that makes bread rise: carbon dioxide; What happens to the other product: ethanol evaporates during baking
B.Product that makes bread rise: carbon dioxide; What happens to the other product: lactic acid is broken down by heat
C.Product that makes bread rise: oxygen; What happens to the other product: carbon dioxide dissolves in the dough
D.Product that makes bread rise: ethanol; What happens to the other product: carbon dioxide escapes into the air
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解題
Anaerobic respiration in yeast (alcohol fermentation) produces carbon dioxide and ethanol. Carbon dioxide gas is trapped as bubbles in the dough, causing it to expand and rise. The other major product, ethanol, is volatile and evaporates during the baking process because of the high oven temperatures.
評分準則
1 mark for selecting correct option A. Correctly identifies carbon dioxide as the gas causing dough expansion and ethanol as the evaporating byproduct.
題目 31 · multiple_choice
1 分
Which statement about the Human Immunodeficiency Virus (HIV) is correct?
A.HIV is a bacterium that can be treated successfully with standard courses of antibiotics.
B.HIV causes an increase in the number of platelets, which leads to blood clots in organs.
C.HIV destroys lymphocytes, which reduces the body's ability to produce antibodies.
D.HIV is transmitted through the air by droplets when an infected person coughs or sneezes.
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解題
HIV is a virus that specifically targets and destroys lymphocytes (helper T-cells) in the human immune system. Because lymphocytes are responsible for antibody production and regulating the immune response, their destruction weakens the body's immune system, eventually leading to AIDS. HIV cannot be treated with antibiotics (which target bacteria), does not increase platelet counts, and is not transmitted via airborne droplets.
評分準則
1 mark for selecting correct option C. Correctly identifies that HIV destroys lymphocytes, leading to a decreased capacity to produce antibodies.
題目 32 · multiple_choice
1 分
A student takes a micrograph of a plant cell. The actual length of the cell is \(0.08\text{ mm}\). In the micrograph, the image of the cell measures \(40\text{ mm}\) in length. What is the magnification of the image?
A.\(\times 0.002\)
B.\(\times 50\)
C.\(\times 500\)
D.\(\times 5000\)
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解題
The formula for magnification is \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Here, the image size is \(40\text{ mm}\) and the actual size is \(0.08\text{ mm}\). Substituting these values into the equation gives: \(\text{Magnification} = \frac{40}{0.08} = 500\). Thus, the magnification is \(\times 500\).
評分準則
1 mark for selecting correct option C. Correctly applies the magnification formula and divides image size by actual size to get 500.
題目 33 · 選擇題
1 分
Four similar-sized pieces of potato cylinder were placed in sucrose solutions of different concentrations: 0.0 mol/dm³, 0.2 mol/dm³, 0.6 mol/dm³, and 1.0 mol/dm³. After two hours, the change in mass of each potato cylinder was measured. Which concentration of sucrose solution would cause the greatest percentage decrease in the mass of the potato cylinder?
A.0.0 mol/dm³
B.0.2 mol/dm³
C.0.6 mol/dm³
D.1.0 mol/dm³
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解題
At 1.0 mol/dm³, the external solution has the lowest water potential (highest solute concentration) compared to the potato cells. Water will move out of the potato cells by osmosis down a water potential gradient at the fastest rate, resulting in the greatest percentage decrease in mass.
評分準則
1 mark: correct option D selected.
題目 34 · 選擇題
1 分
In humans, the allele for wet earwax (W) is dominant to the allele for dry earwax (w). A man who is heterozygous for wet earwax and a woman who has dry earwax have children. What is the probability that their first child will have dry earwax?
A.0%
B.25%
C.50%
D.75%
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解題
The man's genotype is Ww (heterozygous) and the woman's genotype is ww (recessive homozygous). A cross between Ww and ww yields 50% Ww (wet earwax) and 50% ww (dry earwax) offspring. Thus, the probability of dry earwax is 50%.
評分準則
1 mark: correct option C selected.
題目 35 · 選擇題
1 分
A potometer is used to measure the rate of water uptake of a leafy shoot. Which set of environmental conditions would result in the lowest rate of water uptake?
A.High humidity, low temperature, and still air
B.Low humidity, high temperature, and moving air
C.High humidity, high temperature, and still air
D.Low humidity, low temperature, and moving air
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解題
Transpiration rate is lowest under conditions of high relative humidity, low temperature, and still air (no wind). High humidity reduces the water potential gradient between the leaf interior and the air; low temperature decreases the kinetic energy of water molecules; and still air allows water vapour to accumulate near stomata, reducing evaporation.
評分準則
1 mark: correct option A selected.
題目 36 · 選擇題
1 分
Why is the enzyme pectinase used during the industrial production of fruit juices?
A.To break down the plant cell walls, increasing the juice yield and making the juice clearer.
B.To convert glucose into sucrose to make the juice taste sweeter.
C.To digest bacterial cell walls to sterilise the juice and prevent spoilage.
D.To ferment sugars into alcohol to produce a carbonated beverage.
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解題
Pectinase breaks down pectin, which is a key component of plant cell walls. Breaking down pectin weakens the cell structures to release more juice (increasing yield) and degrades suspended particles, making the juice clear rather than cloudy.
評分準則
1 mark: correct option A selected.
題目 37 · 選擇題
1 分
Which row correctly describes the pathogen that causes HIV/AIDS and its primary effect on the body's immune system?
A.Pathogen: bacterium | Effect on immune system: destroys phagocytes, decreasing antibody production
B.Pathogen: virus | Effect on immune system: destroys lymphocytes, decreasing antibody production
C.Pathogen: bacterium | Effect on immune system: destroys lymphocytes, decreasing engulfing of pathogens
D.Pathogen: virus | Effect on immune system: destroys phagocytes, decreasing engulfing of pathogens
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解題
HIV (Human Immunodeficiency Virus) is a virus. It primarily attacks and destroys lymphocytes (specifically helper T cells), which coordinates the immune response. A reduction in functioning lymphocytes directly decreases the body's ability to produce antibodies.
評分準則
1 mark: correct option B selected.
題目 38 · 選擇題
1 分
The diagram outlines the pathway of a reflex arc when a hand touches a sharp pin:
Stimulus -> Receptor -> Neurone X -> Neurone Y -> Neurone Z -> Effector -> Response
Which row correctly identifies Neurone X, Neurone Y, and Neurone Z?
In a reflex arc, Neurone X is the sensory neurone, which transmits impulses from the receptor to the central nervous system. Neurone Y is the relay neurone in the spinal cord, and Neurone Z is the motor neurone, which carries the impulse to the effector.
評分準則
1 mark: correct option C selected.
題目 39 · 選擇題
1 分
A child receives a vaccine containing weakened pathogens. Which statement correctly describes the type of immunity produced and how it is achieved?
A.Active immunity, because the body produces its own antibodies and memory cells in response to the antigens.
B.Passive immunity, because antibodies are injected directly into the body.
C.Active immunity, because ready-made antibodies are transferred from an external source.
D.Passive immunity, because the body does not need to produce its own memory cells.
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解題
Vaccination with a weakened pathogen triggers an active immune response because the host's own immune system is stimulated to produce antibodies and memory cells against the introduced antigens. This provides long-term protection.
評分準則
1 mark: correct option A selected.
題目 40 · 選擇題
1 分
Which row correctly identifies the structural features and main functions of xylem vessels?
A.Structure: made of living cells with cytoplasm | Function: transports sucrose and amino acids
B.Structure: made of dead cells with lignified walls | Function: transports water and mineral ions
C.Structure: made of living cells with sieve plates | Function: transports water and mineral ions
D.Structure: made of dead cells with lignified walls | Function: transports sucrose and amino acids
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解題
Xylem vessels consist of dead cells aligned end-to-end to form a continuous hollow tube. Their walls are strengthened with lignin. Their primary role is the transport of water and dissolved minerals from the roots to the leaves.
評分準則
1 mark: correct option B selected.
Paper 4: Theory (Extended)
Answer all questions. Write your answers in the spaces provided on the question paper. You should show all your working and use appropriate units.
6 題目 · 79.98 分
題目 1 · Structured Extended Theory
13.33 分
A student investigated the effect of different concentrations of sucrose solution on the mass of potato tissue cylinders. Six potato cylinders of equal length and diameter were weighed, placed in different concentrations of sucrose solution for 2 hours, and then reweighed. (a) Define the term osmosis. [3] (b) The potato cylinder placed in 0.0 mol/dm\(^{3}\) sucrose solution showed a significant increase in mass. Explain, in terms of water potential, why this increase occurred. [4] (c) A potato cylinder had an initial mass of 4.20 g. After 2 hours in a 0.8 mol/dm\(^{3}\) sucrose solution, its mass was 3.57 g. Calculate the percentage change in mass of this potato cylinder. Show your working. [3] (d) Describe the appearance of cells from the potato cylinder placed in the 0.8 mol/dm\(^{3}\) sucrose solution and explain this appearance using the term plasmolysis. [3.33]
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解題
(a) Osmosis is the net movement of water molecules from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution), down a water potential gradient, through a partially permeable membrane. (b) The water potential of the 0.0 mol/dm\(^{3}\) sucrose solution is higher than the water potential inside the potato cell vacuoles/cytoplasm. Water molecules moved into the cells by osmosis, down a water potential gradient, across the partially permeable cell membrane. This entry of water increased the turgor pressure of the cells, causing them to expand and the overall mass of the potato cylinder to increase. (c) Change in mass = 3.57 g - 4.20 g = -0.63 g. Percentage change = (-0.63 / 4.20) * 100 = -15.0%. (d) The cells would appear plasmolysed. The cell vacuole shrinks and the cytoplasm pulls away from the cell wall, leaving a visible gap. This happens because the sucrose solution has a lower water potential than the cell sap, causing water to leave the vacuole by osmosis down a water potential gradient.
評分準則
(a) Max 3 marks: net movement of water molecules [1]; from higher water potential to lower water potential / down a water potential gradient [1]; through a partially permeable membrane [1]. (b) Max 4 marks: water potential of solution is higher than inside cells [1]; water enters by osmosis [1]; across a partially permeable membrane [1]; cells become turgid / gain mass [1]. (c) Max 3 marks: correct calculation of mass change (-0.63 g) [1]; division of change by initial mass (-0.63 / 4.20) [1]; correct percentage value (-15% or 15% decrease) [1]. (d) Max 3.33 marks: cell membrane/cytoplasm pulls away from the cell wall [1]; vacuole shrinks / volume of cell sap decreases [1]; water left cell by osmosis [1]; cells described as plasmolysed / flaccid [0.33].
題目 2 · Structured Extended Theory
13.33 分
Immunity can be acquired either actively or passively. (a) Distinguish between active immunity and passive immunity, including how each is acquired and how long the protection lasts. [4] (b) Explain how the process of vaccination stimulates active immunity to protect a person against a specific disease, such as measles. [5] (c) Antibodies are proteins produced by lymphocytes. Explain how antibodies defend the body against pathogens. [4.33]
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解題
(a) Active immunity is the defense against a pathogen by antibody production in the body, which is long-term. It is acquired via natural infection or vaccination. Passive immunity is temporary, short-term defense using antibodies from another organism (e.g., across placenta/breast milk or by injection). It does not produce memory cells. (b) A vaccine contains harmless, weakened, or dead pathogens, or just their antigens. Once injected, these antigens are detected by lymphocytes. Specific lymphocytes divide by mitosis to clone themselves and produce complementary antibodies. Memory cells are also produced. If the live pathogen is encountered in the future, memory cells recognize the antigens and rapidly produce a high concentration of antibodies, destroying the pathogen before symptoms appear. (c) Antibodies have specific shapes complementary to antigens on pathogens. They bind to antigens, causing agglutination (clumping of pathogens), which prevents their spread and makes phagocytosis by phagocytes easier. They can also act as markers for phagocytosis, directly cause lysis of the pathogen, or neutralize toxins produced by the pathogen.
評分準則
(a) Max 4 marks: active involves body's own antibody production vs passive involves receiving external antibodies [1]; active produces memory cells vs passive does not [1]; active is long-term vs passive is short-term/temporary [1]; active acquired by infection/vaccine vs passive acquired by placenta/breast milk/injection [1]. (b) Max 5 marks: vaccine contains dead/weakened pathogen or antigens [1]; lymphocytes recognize antigens [1]; lymphocytes divide by mitosis [1]; antibodies produced [1]; memory cells formed [1]; rapid/large antibody response on future infection [1]. (c) Max 4.33 marks: antibodies are complementary to specific antigens [1]; bind to antigens [1]; cause agglutination / clumping [1]; make phagocytosis easier [1]; neutralize toxins [0.33].
題目 3 · Structured Extended Theory
13.33 分
Albinism in humans is an inherited condition characterized by the absence of pigment in the skin, hair, and eyes. It is caused by a recessive allele, \(a\). The dominant allele, \(A\), leads to normal pigment production. (a) Define the terms: (i) genotype [1] (ii) phenotype [2] (b) A man and a woman who both have normal skin pigmentation but are heterozygous for albinism plan to have a child. Complete a genetic diagram to show: the genotypes of the parents, the genotypes of the gametes, the possible genotypes of the offspring, and the probability of the child having albinism. [5] (c) Explain why this heterozygous couple could have four children, none of whom have albinism. [3] (d) State the term used to describe an individual who has two different alleles for a particular gene, and state the term for an individual who has two of the same alleles. [2.33]
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解題
(a) (i) Genotype is the genetic make-up of an organism in terms of the alleles present. (ii) Phenotype is the observable features of an organism, resulting from its genotype and its environment. (b) Parent genotypes: \(Aa\) and \(Aa\). Gametes: \(A\) and \(a\) from both parents. Offspring genotypes: \(AA\) (normal), \(Aa\) (normal), \(Aa\) (normal), \(aa\) (albinism). Offspring ratio: 3 normal : 1 albinism. Probability of albinism: 25% or 0.25 or 1/4. (c) Each fertilization is a random, independent event. There is a 75% probability for each child to inherit at least one dominant allele (normal pigment). Probability does not guarantee exact ratios in small sample sizes; the outcome of previous children does not affect future ones. (d) An individual with two different alleles is heterozygous. An individual with two identical alleles is homozygous.
評分準則
(a) (i) genotype: alleles present in an organism [1]. (ii) phenotype: observable features [1]; influence of genotype and environment [1]. (b) Max 5 marks: parent genotypes Aa x Aa [1]; gametes A, a and A, a shown separately [1]; Punnett square with correct genotypes (AA, Aa, Aa, aa) [1]; phenotypes linked to genotypes [1]; probability of albinism 25% or 1/4 [1]. (c) Max 3 marks: fertilization is a random / chance event [1]; each child has 75% chance of normal pigment [1]; independent events / previous births do not affect future births [1]. (d) Max 2.33 marks: heterozygous for different alleles [1]; homozygous for same alleles [1]; both correct [0.33].
題目 4 · Structured Extended Theory
13.33 分
Yeast, *Saccharomyces cerevisiae*, is a single-celled fungus widely used in biotechnology. (a) Describe how yeast is used in the bread-making process and explain the biological reasons for the changes that occur in the dough. [4] (b) Yeast can also respire anaerobically to produce ethanol, which is used as a biofuel. Write the balanced chemical equation for this process. [3] (c) Discuss the environmental and economic advantages and disadvantages of using biofuels, such as ethanol from yeast fermentation, compared to fossil fuels. [6.33]
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解題
(a) Yeast is mixed with flour, water, and sugar to make dough. Yeast first respires aerobically, then as oxygen is depleted, it respires anaerobically. This anaerobic respiration (fermentation) produces carbon dioxide gas and ethanol. The carbon dioxide gas is trapped in the dough, forming bubbles that cause the dough to rise. When baked, the high temperature kills the yeast and evaporates the ethanol, leaving a spongy bread structure. (b) \(C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2\). (c) Advantages: Biofuels are renewable; they are carbon-neutral because the plants grown for fuel absorb carbon dioxide via photosynthesis, offsetting emissions from combustion; they reduce dependence on finite fossil fuels. Disadvantages: Large areas of arable land are needed to grow crops, causing deforestation, habitat loss, and reduction in biodiversity; crop production for fuel reduces food supply, increasing food prices; fermentation and distillation require energy, which often still comes from fossil fuels.
評分準則
(a) Max 4 marks: yeast mixed with flour/water/sugar [1]; respires anaerobically to produce carbon dioxide [1]; gas bubbles trapped causing dough to rise [1]; baking kills yeast / evaporates ethanol [1]. (b) Max 3 marks: glucose reactant formula correct (\(C_6H_{12}O_6\)) [1]; products formulas correct (\(C_2H_5OH\) and \(CO_2\)) [1]; balanced equation with 2 in front of ethanol and carbon dioxide [1]. (c) Max 6.33 marks: Advantages (max 3): renewable [1]; carbon-neutral [1]; conserves fossil fuel reserves [1]. Disadvantages (max 3): land use/deforestation/biodiversity loss [1]; food vs fuel competition / high food prices [1]; processing requires energy [1]; total combined score max 6.33.
題目 5 · Structured Extended Theory
13.33 分
Reflex actions protect the body from harm by allowing rapid, involuntary responses to stimuli. (a) Describe the pathway of a nerve impulse in a reflex arc when a person accidentally touches a hot stove. Name all the neurones, receptor, and effector involved in this pathway. [5] (b) A junction between two neurones is called a synapse. Explain how a nerve impulse is transmitted across a synapse from one neurone to the next. [5] (c) (i) Define the term synapse. [1] (ii) Explain why neurotransmitters ensure that nerve impulses can only travel in one direction. [2.33]
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解題
(a) The stimulus (heat) is detected by temperature receptors in the skin. An electrical impulse is generated and travels along the sensory neurone to the spinal cord (CNS). In the gray matter of the spinal cord, the impulse crosses a synapse to a relay neurone. It then crosses another synapse to a motor neurone. The impulse travels along the motor neurone to the effector, which is a muscle in the arm. The muscle contracts, pulling the hand away. (b) When an electrical impulse reaches the pre-synaptic membrane of the axon terminal, it causes vesicles containing neurotransmitters to move to and fuse with the membrane. The neurotransmitters are released into the synaptic cleft by exocytosis. They diffuse across the gap and bind to specific receptor proteins on the post-synaptic membrane. This binding triggers a new electrical impulse in the post-synaptic neurone. (c) (i) A synapse is a junction between two neurones. (ii) Neurotransmitter molecules are only stored in vesicles within the pre-synaptic neurone's synaptic bulb. Also, the specific receptor proteins are only located on the post-synaptic membrane. Therefore, neurotransmitters can only diffuse down their concentration gradient from the pre-synaptic side to the post-synaptic side, ensuring one-way transmission.
評分準則
(a) Max 5 marks: receptor detects heat and generates impulse [1]; impulse travels along sensory neurone to CNS [1]; passes across synapse to relay neurone in spinal cord [1]; passes across synapse to motor neurone [1]; travels to effector/muscle causing contraction [1]. (b) Max 5 marks: impulse arrives at pre-synaptic membrane [1]; vesicles fuse with membrane [1]; neurotransmitter released into synaptic cleft [1]; neurotransmitter diffuses across cleft [1]; binds to specific receptors on post-synaptic membrane [1]. (c) (i) junction between two neurones [1]. (ii) Max 2.33 marks: vesicles containing neurotransmitters only in pre-synaptic terminal [1]; receptors only on post-synaptic membrane [1]; hence diffusion is unidirectional [0.33].
題目 6 · Structured Extended Theory
13.33 分
Vascular bundles in plants contain xylem and phloem tissues, which are adapted for transport and support. (a) Describe how the structure of xylem vessels is adapted to perform its functions of transporting water and supporting the plant. [5] (b) Compare the functions of xylem and phloem. In your answer, refer to the substances transported, the direction of movement, and the energy requirements of each tissue. [5] (c) State the distribution of xylem and phloem inside: (i) a dicotyledonous stem [2] (ii) a dicotyledonous root [1.33]
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解題
(a) Xylem vessels consist of dead, hollow cells joined end-to-end to form a continuous tube with no end walls, allowing water to flow with minimal resistance. Their cell walls are thickened with lignin, a strong, waterproof substance that prevents the vessels from collapsing under the negative pressure (tension) of transpiration pull. Lignin also provides structural support to help the plant stay upright. The presence of tiny unlignified areas (pits) allows lateral movement of water to neighboring tissues. (b) Xylem transports water and dissolved mineral ions, whereas phloem transports sucrose and amino acids. Transport in xylem is unidirectional, moving only upwards from roots to leaves, while transport in phloem (translocation) is bidirectional, moving from sources (where made) to sinks (where used/stored). Xylem transport is a passive process driven by transpiration pull, while phloem transport is an active process requiring energy from ATP (provided by companion cells). Xylem cells are dead, while phloem cells are living. (c) (i) In a dicotyledonous stem, vascular bundles are arranged in a ring near the outer edge of the stem; within each bundle, xylem is on the inside and phloem is on the outside. (ii) In a dicotyledonous root, the vascular bundle is centrally located; the xylem forms a central star or 'X' shape, and phloem is located in the pockets between the arms of the star.
評分準則
(a) Max 5 marks: continuous, hollow tubes with no end walls/cytoplasm [1]; reduces resistance to water flow [1]; lignified walls prevent collapse under tension [1]; lignin provides structural support [1]; pits allow lateral water movement [1]. (b) Max 5 marks: xylem: water/minerals vs phloem: sucrose/amino acids [1]; xylem: upwards vs phloem: bidirectional [1]; xylem: passive (transpiration pull) vs phloem: active (requires respiration/ATP) [1]; xylem: dead cells vs phloem: living cells (sieve tubes/companion cells) [1]; comparison of source and sink [1]. (c) (i) Max 2 marks: arranged in a ring near the edge [1]; xylem on the inside and phloem on the outside of the bundle [1]. (ii) Max 1.33 marks: vascular bundle is central [1]; xylem forms central star/X shape with phloem between arms [0.33].
Paper 6: Alternative to Practical
Answer all questions. Write your answers in the spaces provided on the question paper. You may use a calculator.
3 題目 · 39.99 分
題目 1 · Experimental and Practical Analysis
13.33 分
A student investigated the effect of different sodium chloride (salt) concentrations on the mass of potato strips.
The student: 1. Cut five potato strips to approximately 50 mm in length. 2. Measured and recorded the initial mass of each strip. 3. Placed each strip into a test-tube containing 20 cm³ of salt solution at a different concentration: 0.0, 0.2, 0.4, 0.6, and 0.8 mol/dm³. 4. Left the strips in the solutions for 45 minutes. 5. Removed the strips, patted them dry with a paper towel, and measured the final mass.
The student's raw results are shown below: - 0.0 mol/dm³: initial mass = 4.20 g; final mass = 4.62 g - 0.2 mol/dm³: initial mass = 4.15 g; final mass = 4.31 g - 0.4 mol/dm³: initial mass = 4.30 g; final mass = 4.13 g - 0.6 mol/dm³: initial mass = 4.05 g; final mass = 3.65 g - 0.8 mol/dm³: initial mass = 4.10 g; final mass = 3.48 g
(a) Prepare a single table to present all the results. Your table should include: - concentration of salt solution (mol/dm³) - initial mass (g) - final mass (g) - change in mass (g) - percentage change in mass (%) Calculate the change in mass and percentage change in mass for the first four concentrations and enter them into your table (do not calculate for 0.8 mol/dm³ yet). [4 marks]
(b) Calculate the percentage change in mass for the potato strip placed in the 0.8 mol/dm³ salt solution. Show your working and give your answer to one decimal place. [2 marks]
(c) Identify: (i) the independent variable. [1 mark] (ii) two variables that should be controlled. [2 marks]
(d) Explain why the student patted the potato strips dry with a paper towel before measuring their final mass. [1.33 marks]
(e) Suggest two improvements to this experiment that would make the results more reliable. [3 marks]
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解題
(a) Table of results: Concentration of salt solution / mol/dm³ | Initial mass / g | Final mass / g | Change in mass / g | Percentage change in mass / % - 0.0 | 4.20 | 4.62 | +0.42 | +10.0 - 0.2 | 4.15 | 4.31 | +0.16 | +3.9 - 0.4 | 4.30 | 4.13 | -0.17 | -4.0 - 0.6 | 4.05 | 3.65 | -0.40 | -9.9
(b) Working: Change in mass = \(3.48 \text{ g} - 4.10 \text{ g} = -0.62 \text{ g}\) Percentage change = \((\frac{-0.62}{4.10}) \times 100 = -15.122\%\) Answer to one decimal place = \(-15.1\%\) (or a decrease of \(15.1\%\))
(c)(i) Concentration of sodium chloride/salt solution. (c)(ii) Any two from: Volume of salt solution (20 cm³); time strips are left in the solutions (45 minutes); temperature; variety/type of potato; initial dimensions/surface area of potato strips.
(d) To remove any excess liquid/surface water on the outside of the potato strips, which would add to the measured final mass and lead to an inaccurate calculation of mass change.
(e) Any two from: Repeat the experiment at each concentration and calculate a mean; cover the test tubes with bungs/foil to prevent evaporation; use a cork borer to cut cylinders of identical diameter/surface area; use an electronic balance with a higher precision (e.g. 3 decimal places).
評分準則
(a) [4 marks] - 1 mark for structured table layout with clear, bordered columns and rows. - 1 mark for fully complete, correct headers with units (e.g. Concentration / mol/dm³, Mass / g, % change). - 1 mark for calculating correct change in mass for 0.0 to 0.6 mol/dm³ (must show sign: +0.42, +0.16, -0.17, -0.40). - 1 mark for correct calculations of percentage change to 1 decimal place (+10.0%, +3.9%, -4.0%, -9.9%).
(b) [2 marks] - 1 mark for correct working shown: \(\frac{3.48 - 4.10}{4.10} \times 100\) (or \(\frac{-0.62}{4.10} \times 100\)). - 1 mark for final correct answer of \(-15.1\%\) (accept a decrease of \(15.1\%\); reject without minus sign or without indicating decrease).
(c) [3 marks total] - (i) 1 mark for correctly identifying the independent variable as concentration of salt solution. - (ii) 2 marks for any two control variables (1 mark each). Accept: volume of solution, duration of soaking, temperature, potato source, surface area/diameter.
(d) [1.33 marks] - 1 mark for stating that it removes excess/clinging liquid on the surface. - 0.33 marks for explaining this prevents a false increase in measured final mass/inaccuracy in calculated change.
(e) [3 marks] - 1 mark each for up to two valid improvements (max 2 marks). - 1 mark for linking an improvement to increasing reliability (e.g. 'repeating allows identification of anomalies and calculation of a more reliable mean').
題目 2 · Experimental and Practical Analysis
13.33 分
A student investigated the transpiration rate of a leafy shoot using a simple potometer. They recorded the distance moved by an air bubble along a capillary tube in 10 minutes under three different conditions: - Condition A: Still air (control) - Condition B: Moving air (fan) - Condition C: Humid air (enclosed in a clear plastic bag)
The student's raw data (distance moved by the bubble in 10 minutes in mm) is recorded in three replicates for each condition: - Condition A: 12 mm, 14 mm, 13 mm (Mean = 13.0 mm) - Condition B: 32 mm, 36 mm, 34 mm (Mean = 34.0 mm) - Condition C: 4 mm, 3 mm, 5 mm (Mean = 4.0 mm)
(a) Plot a bar chart to show the mean distance moved by the air bubble in 10 minutes for the three conditions. [5 marks]
(b) Explain the difference in the rate of bubble movement between moving air (Condition B) and humid air (Condition C) by referencing water vapor concentration gradients. [3.33 marks]
(c) The actual width of the stem of the plant used in this experiment is 2.5 mm. In a photomicrograph of this stem, the measured width is 75 mm. Calculate the magnification of the photomicrograph. Show your working. [2 marks]
(d) Outline the practical procedure to prepare a temporary microscope slide of a thin stem section to observe the xylem vessels under a light microscope. [3 marks]
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解題
(a) Graph description: - x-axis: labeled 'Condition' (A: Still air, B: Moving air, C: Humid air) - y-axis: labeled 'Mean distance moved by air bubble / mm' - Linear, sensible scale on y-axis (e.g., 0 to 40 mm, with increments of 5 or 10) - Three separate bars plotted with spaces between them, matching values of 13.0 mm, 34.0 mm, and 4.0 mm. - Bars are equal in width and plotted accurately using a ruler.
(b) Moving air (Condition B) removes the accumulated water vapor from around the stomata of the leaves, maintaining or steepening the water vapor concentration gradient between the inside of the leaf and the atmosphere. This increases the rate of diffusion of water vapor (transpiration), causing the bubble to move faster. In contrast, humid air (Condition C) increases the water vapor concentration around the leaf, reducing the water vapor concentration gradient between the inside of the leaf and the outside air. This slows down the rate of diffusion (transpiration), causing the bubble to move slower.
(d) Practical procedure: 1. Use a sharp razor blade or scalpel to cut extremely thin transverse sections of the stem. 2. Place the thinnest section onto the center of a clean glass microscope slide. 3. Add a few drops of a suitable stain (e.g. toluidine blue or iodine) to color the xylem walls / add water. 4. Gently lower a glass coverslip at an angle over the specimen using a mounted needle to avoid trapping air bubbles.
評分準則
(a) [5 marks] - A (Axes): Both axes fully labeled with units (x-axis: Condition, y-axis: Mean distance moved / mm). - S (Scale): Linear y-axis scale, starting at 0, covering more than half the grid. - P (Plotting): All 3 bars plotted accurately to within half a small grid square. - B (Bar characteristics): Bars drawn with a ruler, distinct with gaps between them, and of equal width. - L (Line/Clarity): Neat presentation with no untidy scribbles.
(b) [3.33 marks] - 1 mark for stating that moving air (Condition B) maintains/steepens the water vapor concentration gradient by blowing water vapor away. - 1 mark for stating that humid air (Condition C) reduces the water vapor concentration gradient by keeping water vapor around the leaf. - 1.33 marks for linking a steeper gradient to faster diffusion/transpiration and a shallower gradient to slower diffusion/transpiration.
(c) [2 marks] - 1 mark for correct working shown: \(\frac{75}{2.5}\). - 1 mark for correct magnification value of \(\times 30\) or \(30\) (do not accept units like mm here).
(d) [3 marks] - 1 mark for cutting thin section(s) with a sharp blade/scalpel. - 1 mark for staining the specimen / placing it on a slide with a liquid/stain. - 1 mark for lowering the coverslip carefully at an angle / avoiding air bubbles.
題目 3 · Experimental and Practical Analysis
13.33 分
A student investigated the effect of temperature on the rate of anaerobic respiration in yeast. They used a gas syringe to measure the volume of carbon dioxide produced by a yeast-glucose mixture over 10 minutes at five different temperatures: 20 °C, 30 °C, 40 °C, 50 °C, and 60 °C.
Their results are shown in the data below: - 20 °C: 6.2 cm³ - 30 °C: 14.5 cm³ - 40 °C: 22.8 cm³ - 50 °C: 11.2 cm³ - 60 °C: 1.5 cm³
(a) Draw a line graph on grid paper to show the relationship between temperature and the volume of carbon dioxide produced in 10 minutes. [5 marks]
(b) (i) Describe the trend shown by the graph. [2 marks] (ii) Explain the results at 40 °C and 60 °C using enzyme theory. [3.33 marks]
(c) State how the student could modify this procedure to determine the optimum temperature for yeast respiration more accurately. [1 mark]
(d) Explain why a layer of liquid paraffin (oil) is sometimes floated on top of the yeast-glucose mixture in experiments of this type. [2 marks]
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解題
(a) Graph description: - x-axis: Temperature / °C (labeled and scaled, e.g. 0 to 70 with 10-degree increments) - y-axis: Volume of carbon dioxide / cm³ (labeled and scaled, e.g. 0 to 25 with 5-unit increments) - Points plotted accurately at (20, 6.2), (30, 14.5), (40, 22.8), (50, 11.2), and (60, 1.5). - Points joined with a neat, smooth curve or straight lines from point to point. - No extrapolation beyond the plotted data points.
(b)(i) As temperature increases from 20 °C to 40 °C, the volume of carbon dioxide produced increases, reaching a peak of 22.8 cm³ at 40 °C. Above 40 °C, as the temperature increases to 60 °C, the volume of carbon dioxide decreases rapidly. (b)(ii) At 40 °C, enzymes in the yeast have high kinetic energy, leading to a high frequency of successful collisions between active sites and substrate molecules, resulting in a high rate of anaerobic respiration. At 60 °C, the temperature is too high, which denatures the enzymes. The shape of the active sites changes so they are no longer complementary to the substrate, stopping respiration.
(c) Repeat the experiment at smaller temperature intervals between 30 °C and 50 °C (e.g., every 2 °C or 5 °C).
(d) The layer of liquid paraffin blocks oxygen from entering the yeast-glucose mixture from the air. This ensures anaerobic conditions are maintained, so that only anaerobic respiration occurs.
評分準則
(a) [5 marks] - A (Axes): Both axes labeled with units (x-axis: Temperature / °C, y-axis: Volume of carbon dioxide / cm³). - S (Scale): Linear scale starting at origin, using more than half of the grid area. - P (Plotting): All 5 points plotted correctly to within half a small square. - L (Line): Points joined cleanly by straight lines point-to-point or a smooth curve; no double lines. - K (Key/No extrapolation): No lines extended past 60 °C or below 20 °C unless drawn to origin properly without assumption.
(b) [5.33 marks total] - (i) [2 marks]: 1 mark for stating that volume of CO2 increases with temperature up to a maximum at 40 °C; 1 mark for stating that it decreases rapidly at higher temperatures (above 40 °C). - (ii) [3.33 marks]: 1 mark for explaining high kinetic energy/more successful collisions at 40 °C; 1 mark for stating that enzymes denature at 60 °C; 1.33 marks for explaining that denaturing changes the shape of the active site so substrate cannot fit.
(c) [1 mark] - 1 mark for suggesting smaller/narrower temperature intervals (between 30 °C and 50 °C). Reject 'do more temperatures' without specifying smaller intervals or a range.
(d) [2 marks] - 1 mark for stating that paraffin prevents oxygen from entering / dissolving in the mixture. - 1 mark for explaining this ensures conditions are anaerobic / prevents aerobic respiration.
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