An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V1) Cambridge IGCSE 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 题目 · 40 分
题目 1 · 選擇題
1 分
An animal cell and a plant cell are placed in distilled water. Which row correctly describes the movement of water and the final state of each cell?
A.animal cell: water enters and cell bursts; plant cell: water enters and cell becomes turgid
B.animal cell: water leaves and cell shrinks; plant cell: water enters and cell becomes turgid
C.animal cell: water enters and cell bursts; plant cell: water leaves and cell becomes flaccid
D.animal cell: water leaves and cell shrinks; plant cell: water leaves and cell becomes flaccid
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解题
Distilled water has a higher water potential than the cytoplasm of both cells. Therefore, water moves into both cells down a water potential gradient by osmosis. The animal cell has no cell wall, so the increased internal pressure causes it to burst. The plant cell has a strong cellulose cell wall which prevents it from bursting, making the cell turgid.
评分标准
1 mark for the correct option A.
题目 2 · 選擇題
1 分
A photomicrograph shows a human cheek cell. The actual diameter of this cheek cell is 60 \(\mu\text{m}\). If the magnification of the image is \(\times 750\), what is the diameter of the cheek cell in the photomicrograph?
A.4.5 mm
B.45 mm
C.450 mm
D.4500 mm
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解题
Using the formula Image size (I) = Actual size (A) \(\times\) Magnification (M): \(I = 60\ \mu\text{m} \times 750 = 45\,000\ \mu\text{m}\). To convert micrometres to millimetres, divide by 1000: \(45\,000\ \mu\text{m} / 1000 = 45\text{ mm}\).
评分标准
1 mark for the correct option B.
题目 3 · 選擇題
1 分
A woman of blood group A has a child with a man of blood group B. The child has blood group O. What is the probability that their next child will have blood group AB?
A.0%
B.25%
C.50%
D.75%
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解题
Since the child is blood group O (genotype \(I^o I^o\)), each parent must have contributed an \(I^o\) allele. Thus, the mother's genotype is \(I^A I^o\) and the father's genotype is \(I^B I^o\). A cross between these genotypes yields offspring with genotypes: \(I^A I^B\) (group AB), \(I^A I^o\) (group A), \(I^B I^o\) (group B), and \(I^o I^o\) (group O) in a 1:1:1:1 ratio. Therefore, the probability of having a child with blood group AB is 1 out of 4, which is 25%.
评分标准
1 mark for the correct option B.
题目 4 · 選擇題
1 分
A student uses a potometer to measure the rate of transpiration of a leafy shoot. Which combination of environmental conditions would result in the slowest movement of the air bubble in the potometer?
A.high temperature, low humidity, high wind speed
B.high temperature, high humidity, low wind speed
C.low temperature, low humidity, high wind speed
D.low temperature, high humidity, still air
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解题
The movement of the air bubble in a potometer represents the rate of transpiration. Transpiration is slowest under conditions of low temperature (slower evaporation), high humidity (reduced concentration gradient of water vapour), and still air (allowing water vapour to accumulate around the stomata, further reducing the diffusion gradient).
评分标准
1 mark for the correct option D.
题目 5 · 選擇題
1 分
Which row correctly describes the type of immunity and the method of acquisition when a person receives an injection of ready-made antibodies?
A.type of immunity: active; method of acquisition: natural
B.type of immunity: active; method of acquisition: artificial
C.type of immunity: passive; method of acquisition: natural
D.type of immunity: passive; method of acquisition: artificial
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解题
Receiving ready-made antibodies represents passive immunity because the individual's own body does not produce the antibodies. Since the antibodies are introduced artificially through an injection, the method of acquisition is artificial.
评分标准
1 mark for the correct option D.
题目 6 · 選擇題
1 分
During a reflex action, in which order does an electrical impulse travel through the neurones?
B.sensory neurone \(\rightarrow\) motor neurone \(\rightarrow\) relay neurone
C.sensory neurone \(\rightarrow\) relay neurone \(\rightarrow\) motor neurone
D.relay neurone \(\rightarrow\) sensory neurone \(\rightarrow\) motor neurone
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解题
In a reflex arc, the receptor detects a stimulus and sends an electrical impulse along the sensory neurone. This neurone synapses with a relay neurone in the central nervous system, which in turn passes the impulse to a motor neurone that carries it to the effector.
评分标准
1 mark for the correct option C.
题目 7 · 選擇題
1 分
Why is a cold-water jacket used around an industrial fermenter?
A.to prevent the temperature from rising too high due to heat released during respiration
B.to maintain a high temperature that denatures the enzymes of competing bacteria
C.to provide a constant source of water for the respiration of the microorganisms
D.to keep the fermenter very cold so that the microorganisms reproduce more slowly
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解题
Respiration by microorganisms in a fermenter is an exothermic process that releases heat. If this heat is not removed, the temperature inside the fermenter will rise too high and denature the enzymes of the microorganisms, stopping the production. A cold-water jacket cools the fermenter to maintain the optimum temperature.
评分标准
1 mark for the correct option A.
题目 8 · 選擇題
1 分
Which hormone stimulates the repair and thickening of the uterus lining and inhibits the secretion of FSH during the first half of the menstrual cycle?
A.FSH
B.LH
C.oestrogen
D.progesterone
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解题
Oestrogen is secreted by the developing follicles in the ovary. It acts on the uterus to stimulate the repair and thickening of the lining after menstruation. It also exerts negative feedback on the pituitary gland, inhibiting the secretion of FSH to prevent further follicle development.
评分标准
1 mark for the correct option C.
题目 9 · multiple_choice
1 分
Equal volumes of a red blood cell suspension were placed into four different test-tubes, each containing a salt solution of a different concentration. After 30 minutes, a sample from each test-tube was examined under a light microscope.
In which salt concentration would the highest percentage of burst cells be observed?
A.$0.0\text{ g/dm}^3$ (pure water)
B.$9.0\text{ g/dm}^3$
C.$18.0\text{ g/dm}^3$
D.$36.0\text{ g/dm}^3$
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解题
Red blood cells are animal cells and do not have a cell wall. When placed in a solution with a higher water potential than their cytoplasm (such as pure water at $0.0\text{ g/dm}^3$), water enters the cells by osmosis down a water potential gradient. This causes the cells to swell and eventually burst (lyse). At higher salt concentrations (such as $9.0\text{ g/dm}^3$ which is approximately isotonic, or hypertonic concentrations like $18.0\text{ g/dm}^3$ and $36.0\text{ g/dm}^3$), water will either remain in dynamic equilibrium or leave the cells, causing them to shrink (crenate) but not burst.
评分标准
Award 1 mark for the correct option (A). - Reject other options as they represent concentrations where cells would either remain intact (B) or shrink due to water loss (C and D).
题目 10 · multiple_choice
1 分
Which row correctly identifies where neurotransmitter molecules are released from, how they move across the synaptic gap, and where their complementary receptors are located?
A.Released from: presynaptic neurone | Movement: active transport | Receptors on: postsynaptic membrane
D.Released from: postsynaptic neurone | Movement: active transport | Receptors on: presynaptic membrane
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解题
Neurotransmitter molecules are stored in vesicles within the presynaptic neurone. When an impulse arrives, they are released into the synaptic gap and travel across it to the postsynaptic membrane by diffusion (down a concentration gradient). They then bind to complementary receptors located on the postsynaptic membrane to initiate a new electrical impulse.
评分标准
Award 1 mark for the correct option (C). - A and D are incorrect because movement is by diffusion, not active transport. - B is incorrect because neurotransmitters are released from the presynaptic neurone and receptors are on the postsynaptic membrane.
题目 11 · multiple_choice
1 分
What is the main biological reason for adding the enzyme pectinase during the commercial production of fruit juice?
A.to break down the cell walls of the fruit, increasing the yield of juice
B.to sterilise the juice by destroying bacterial cell walls
C.to catalyse the synthesis of glucose to make the juice taste sweeter
D.to lower the activation energy of fermentation for alcohol production
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解题
Pectinase is an enzyme that catalyses the breakdown of pectin, a polysaccharide found in plant cell walls. Breaking down pectin weakens the cell walls of the fruit, making it easier to extract the juice (thereby increasing the yield) and helping to clarify the juice by breaking down suspended semi-soluble fibers.
评分标准
Award 1 mark for the correct option (A). - B is incorrect because pectinase does not act as a sterilising agent. - C is incorrect because pectinase breaks down pectin, not starch/sugar to produce glucose directly. - D is incorrect because pectinase is not involved in yeast fermentation.
题目 12 · multiple_choice
1 分
A micrograph of a plant cell shows a chloroplast with an image length of $12\text{ mm}$. The actual length of the chloroplast is $4\ \mu\text{m}$.
What is the magnification of the micrograph?
A.$\times 30$
B.$\times 300$
C.$\times 3000$
D.$\times 30000$
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解题
Use the formula: $\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}$ 1. Convert the image size from millimetres to micrometres: $12\text{ mm} = 12 \times 1000\ \mu\text{m} = 12000\ \mu\text{m}$ 2. Calculate the magnification: $\text{Magnification} = \frac{12000\ \mu\text{m}}{4\ \mu\text{m}} = 3000$ Therefore, the magnification is $\times 3000$.
评分标准
Award 1 mark for the correct option (C). - A, B, and D represent incorrect unit conversions or calculation errors.
题目 13 · multiple_choice
1 分
A man of blood group A and a woman of blood group B have a child. The child is found to have blood group O.
What must be the genotypes of the parents?
A.Father: $I^A I^A$, Mother: $I^B I^B$
B.Father: $I^A I^O$, Mother: $I^B I^B$
C.Father: $I^A I^A$, Mother: $I^B I^O$
D.Father: $I^A I^O$, Mother: $I^B I^O$
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解题
The allele for blood group O ($I^O$) is recessive to both the allele for blood group A ($I^A$) and the allele for blood group B ($I^B$). For a child to have blood group O, they must have the homozygous recessive genotype $I^O I^O$. Therefore, the child must inherit one $I^O$ allele from each parent. Since the father has blood group A, his genotype must be heterozygous ($I^A I^O$). Since the mother has blood group B, her genotype must be heterozygous ($I^B I^O$).
评分标准
Award 1 mark for the correct option (D). - Options A, B, and C are incorrect because at least one parent lacks the recessive $I^O$ allele, making it impossible to produce a child of blood group O ($I^O I^O$).
题目 14 · multiple_choice
1 分
A student uses a potometer to measure the rate of water uptake of a leafy shoot under different environmental conditions.
Which set of conditions will produce the lowest rate of water uptake?
A.Humidity: high | Temperature: low | Wind speed: low
B.Humidity: low | Temperature: high | Wind speed: high
C.Humidity: high | Temperature: high | Wind speed: low
D.Humidity: low | Temperature: low | Wind speed: high
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解题
The rate of water uptake in a potometer is determined by the rate of transpiration. Transpiration is lowest when: - Humidity is high (this decreases the water potential gradient between the inside of the leaf and the external atmosphere). - Temperature is low (this reduces the kinetic energy of water molecules, slowing down evaporation). - Wind speed is low (still air allows a layer of water vapour to accumulate outside the stomata, reducing the water potential gradient). Therefore, the combination of high humidity, low temperature, and low wind speed yields the lowest rate of water uptake.
评分标准
Award 1 mark for the correct option (A). - B, C, and D contain conditions (such as low humidity, high temperature, or high wind speed) that increase the water potential gradient and increase the rate of transpiration.
题目 15 · multiple_choice
1 分
Which statement correctly describes a feature of passive immunity?
A.It provides long-term defence involving the formation of memory cells.
B.It is slow to develop but remains effective in the body for many years.
C.It is achieved when a pathogen stimulates the recipient's own lymphocytes to produce antibodies.
D.It provides fast-acting, short-term protection by receiving ready-made antibodies from another source.
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解题
Passive immunity is the short-term defence against a pathogen by receiving antibodies from an external source (such as from a mother via breast milk/placenta, or via an injection of immunoglobulin antitoxins). Because the recipient's own lymphocytes are not activated, no memory cells are produced, and the protection is only temporary as the foreign antibodies are eventually broken down by the body.
评分标准
Award 1 mark for the correct option (D). - A is incorrect because passive immunity does not involve memory cells. - B is incorrect because long-term protection is a feature of active immunity. - C is incorrect because active immunity, not passive, involves the production of antibodies by the body's own lymphocytes.
题目 16 · multiple_choice
1 分
What is the correct sequence of events leading to the development of antibiotic resistance in a population of bacteria?
A.exposure to antibiotic $\rightarrow$ natural selection of resistant bacteria $\rightarrow$ mutation occurring in response to the antibiotic $\rightarrow$ reproduction of resistant bacteria
B.mutation occurring randomly $\rightarrow$ exposure to antibiotic $\rightarrow$ survival and reproduction of resistant bacteria $\rightarrow$ inheritance of the resistance allele
C.natural selection $\rightarrow$ mutation occurring in response to the antibiotic $\rightarrow$ exposure to antibiotic $\rightarrow$ reproduction of resistant bacteria
D.inheritance of resistance allele $\rightarrow$ mutation occurring randomly $\rightarrow$ exposure to antibiotic $\rightarrow$ survival of all bacteria
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解题
According to the theory of natural selection: 1. A random genetic mutation occurs first in a bacterium, conferring resistance to a specific antibiotic. 2. The bacterial population is then exposed to the antibiotic (which acts as a selective pressure). 3. Non-resistant bacteria are killed, while the mutant resistant bacterium survives. 4. The surviving resistant bacterium reproduces, passing on the advantageous allele to its offspring, increasing its frequency in the population.
评分标准
Award 1 mark for the correct option (B). - A and C are incorrect because mutations occur randomly and are not caused *in response to* the exposure of antibiotics. - D describes an incorrect chronological sequence.
题目 17 · 選擇題
1 分
Four cylinders of plant tissue, each of the same initial length, were placed in different concentrations of sucrose solution for one hour. Cylinder 1 was placed in 0.0 mol dm^-3 solution and its length increased by 5.2%. Cylinder 2 was placed in 0.2 mol dm^-3 solution and its length increased by 1.5%. Cylinder 3 was placed in 0.4 mol dm^-3 solution and its length decreased by 2.1%. Cylinder 4 was placed in 0.6 mol dm^-3 solution and its length decreased by 4.8%. Which statement explains the change in Cylinder 3?
A.Water potential inside Cylinder 3 cells was higher than the sucrose solution, so water entered by osmosis.
B.Water potential inside Cylinder 3 cells was lower than the sucrose solution, so water left by active transport.
C.Water potential inside Cylinder 3 cells was higher than the sucrose solution, so water left by osmosis.
D.Water potential inside Cylinder 3 cells was lower than the sucrose solution, so water entered by active transport.
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解题
The decrease in length of Cylinder 3 indicates that water left the plant cells by osmosis. Water moves from a region of higher water potential (inside the cells) to a region of lower water potential (the 0.4 mol dm^-3 sucrose solution).
评分标准
1 mark for correct option C.
题目 18 · 選擇題
1 分
In a species of plant, flower color is controlled by a codominant gene with two alleles: C^R (red) and C^W (white). Heterozygous plants (C^R C^W) have pink flowers. A cross is made between a pink-flowered plant and a white-flowered plant. What is the expected phenotypic ratio of the offspring?
A.1 red : 1 white
B.1 pink : 1 white
C.1 red : 2 pink : 1 white
D.3 pink : 1 white
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解题
The cross is C^R C^W (pink) x C^W C^W (white). The gametes from the pink plant are C^R and C^W, while the white plant only produces C^W. The offspring genotypes will be 50% C^R C^W (pink) and 50% C^W C^W (white), giving a 1 pink : 1 white phenotypic ratio.
评分标准
1 mark for correct option B.
题目 19 · 選擇題
1 分
A micrograph of a plant cell shows a chloroplast with a length of 12 mm. If the actual length of the chloroplast is 4 micrometres, what is the magnification of the micrograph?
A.x3
B.x300
C.x3000
D.x30000
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解题
Magnification = Image size / Actual size. Image size = 12 mm = 12,000 micrometres. Actual size = 4 micrometres. Magnification = 12,000 / 4 = x3000.
评分标准
1 mark for correct option C.
题目 20 · 選擇題
1 分
Which cell type is specifically targeted and destroyed by the Human Immunodeficiency Virus (HIV), leading to a weakened immune response and the eventual development of AIDS?
A.Red blood cells
B.Phagocytes
C.T-lymphocytes
D.Platelets
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解题
HIV infects and destroys a specific type of white blood cell called helper T-lymphocytes, which play a central role in coordinating the body's immune response.
评分标准
1 mark for correct option C.
题目 21 · 選擇題
1 分
What is the correct sequence of events when an impulse travels across a synapse?
A.neurotransmitter diffuses across cleft -> receptor activation on postsynaptic membrane -> release of neurotransmitter from vesicles
B.release of neurotransmitter from vesicles -> neurotransmitter diffuses across cleft -> receptor activation on postsynaptic membrane
C.receptor activation on postsynaptic membrane -> neurotransmitter diffuses across cleft -> release of neurotransmitter from vesicles
D.neurotransmitter diffuses across cleft -> release of neurotransmitter from vesicles -> receptor activation on postsynaptic membrane
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解题
When an impulse arrives at the synapse, it triggers the release of neurotransmitter chemicals from vesicles in the presynaptic neurone. These neurotransmitters diffuse across the synaptic cleft and bind to specific receptor molecules on the postsynaptic membrane to trigger a new impulse.
评分标准
1 mark for correct option B.
题目 22 · 選擇題
1 分
A child is bitten by a venomous snake and is immediately given an injection of antivenom containing specific antibodies to neutralize the toxin. Which type of immunity does this injection provide?
A.Active, artificial immunity
B.Active, natural immunity
C.Passive, artificial immunity
D.Passive, natural immunity
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解题
Because antibodies are injected directly, the body does not make them itself (passive immunity). Since the antibodies are introduced via a medical treatment (antivenom), it is artificial. Therefore, it provides passive, artificial immunity.
评分标准
1 mark for correct option C.
题目 23 · 選擇題
1 分
A sample of food is tested using three reagents. The Benedict's test gives an orange-red precipitate. The Biuret test results in a blue solution. The Iodine test gives a blue-black color. Which biological molecules are present in this food sample?
A.Reducing sugar and starch only
B.Protein and starch only
C.Reducing sugar and protein only
D.Reducing sugar, protein, and starch
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解题
An orange-red precipitate with Benedict's test indicates reducing sugar is present. A blue color with Biuret test indicates protein is absent (since a purple color would indicate presence). A blue-black color with Iodine test indicates starch is present. Therefore, reducing sugar and starch are present.
评分标准
1 mark for correct option A.
题目 24 · 選擇題
1 分
The diploid chromosome number of a sheep is 54. Which row correctly identifies the type of cell division that produces male gametes in sheep, and the number of chromosomes in each gamete?
A.type of division: meiosis; number of chromosomes: 27
B.type of division: meiosis; number of chromosomes: 54
C.type of division: mitosis; number of chromosomes: 27
D.type of division: mitosis; number of chromosomes: 54
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解题
Gametes (sperm cells) are produced by meiosis, which is a reduction division. The resulting gametes are haploid, meaning they contain half the diploid chromosome number. Therefore, each gamete contains 54 / 2 = 27 chromosomes.
评分标准
1 mark for correct option A.
题目 25 · 選擇題
1 分
A student exercises in a hot environment. Which row correctly describes the changes in the skin blood vessels and the rate of sweating to help lower their body temperature?
A.skin arterioles constrict, sweating increases
B.skin arterioles constrict, sweating decreases
C.skin arterioles dilate, sweating increases
D.skin arterioles dilate, sweating decreases
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解题
In a hot environment, body temperature must be lowered. Skin arterioles dilate (vasodilation) to increase blood flow to capillaries near the skin surface, promoting heat loss by radiation and convection. Sweating also increases, and the evaporation of sweat removes heat energy from the body.
评分标准
Award 1 mark for selecting the correct option C.
题目 26 · 選擇題
1 分
Plant cells with a sap vacuole concentration equivalent to \(0.4\text{ mol dm}^{-3}\) sucrose solution were placed in four test-tubes, each containing a different concentration of sucrose solution. In which solution would the plant cells become most turgid?
A.\(0.1\text{ mol dm}^{-3}\) sucrose solution
B.\(0.4\text{ mol dm}^{-3}\) sucrose solution
C.\(0.6\text{ mol dm}^{-3}\) sucrose solution
D.\(1.0\text{ mol dm}^{-3}\) sucrose solution
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解题
To become turgid, water must enter the plant cells by osmosis. Water moves from a region of higher water potential (less concentrated solution) to a region of lower water potential (more concentrated cell interior). The \(0.1\text{ mol dm}^{-3}\) sucrose solution has the highest water potential compared to the interior of the cells, resulting in the greatest net movement of water into the cells, making them the most turgid.
评分标准
Award 1 mark for selecting the correct option A.
题目 27 · 選擇題
1 分
A photomicrograph of a human red blood cell has a magnification of \(\times 5000\). If the diameter of the red blood cell in the image is \(35\text{ mm}\), what is the actual diameter of the cell?
A.\(0.007\text{ }\mu\text{m}\)
B.\(0.14\text{ }\mu\text{m}\)
C.\(7.0\text{ }\mu\text{m}\)
D.\(142\text{ }\mu\text{m}\)
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解题
Convert the image size of \(35\text{ mm}\) to micrometres (\(\mu\text{m}\)) by multiplying by \(1000\), giving \(35000\text{ }\mu\text{m}\). Using the formula: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}} = \frac{35000\text{ }\mu\text{m}}{5000} = 7\text{ }\mu\text{m}\).
评分标准
Award 1 mark for correct calculation and option C.
题目 28 · 選擇題
1 分
Cystic fibrosis is an inherited condition caused by a recessive allele, \(\text{f}\). Parents who are heterozygous for the condition have a child. What is the probability that their child will be a carrier of cystic fibrosis but not have the disease?
A.\(0.00\)
B.\(0.25\)
C.\(0.50\)
D.\(0.75\)
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解题
Heterozygous parents both have the genotype \(\text{Ff}\). A genetic cross of \(\text{Ff} \times \text{Ff}\) shows offspring genotypes of: \(25\%\text{ FF}\) (normal), \(50\%\text{ Ff}\) (carrier), and \(25\%\text{ ff}\) (affected). Carriers are heterozygous and do not display symptoms of the disease. Therefore, the probability of the child being a carrier is \(0.50\) (or \(50\%\)).
评分标准
Award 1 mark for identifying the carrier probability of 0.50, option C.
题目 29 · 選擇題
1 分
Which row correctly distinguishes between active immunity and passive immunity?
A.active: antibodies are produced by the body itself | passive: memory cells are produced
B.active: antibodies are produced by the body itself | passive: antibodies are introduced from an external source
C.active: antibodies are introduced from an external source | passive: antibodies are produced by the body itself
D.active: memory cells are produced | passive: memory cells are produced
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解题
Active immunity is the defense against a pathogen by antibody production in the body itself, which creates memory cells for long-term protection. Passive immunity is the short-term defense acquired by the introduction of antibodies from an external source (e.g., breast milk or injection), and does not produce memory cells.
评分标准
Award 1 mark for identifying the correct differences between active and passive immunity, option B.
题目 30 · 選擇題
1 分
A student wants to investigate the rate of transpiration from a leafy shoot using a potometer. Which combination of environmental factors would result in the highest rate of transpiration?
A.high humidity, low temperature, strong wind
B.high humidity, high temperature, no wind
C.low humidity, high temperature, strong wind
D.low humidity, low temperature, no wind
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解题
The transpiration rate increases when the concentration gradient of water vapour between the inside of the leaf and the surrounding air is steep. This gradient is increased by high temperatures (which increases evaporation), low humidity (dryer air outside), and strong wind (which removes the saturated water vapour layer accumulating outside the stomata).
评分标准
Award 1 mark for identifying the conditions that maximize the transpiration rate, option C.
题目 31 · 選擇題
1 分
During the industrial production of penicillin in a fermenter, why is sterile air continually bubbled through the liquid medium?
A.to supply carbon dioxide for photosynthesis and mix the contents
B.to supply oxygen for aerobic respiration and keep the fungus in suspension
C.to cool down the fermenter and provide nitrogen for protein synthesis
D.to increase the pH of the mixture and kill contaminating bacteria
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解题
The fungus Penicillium requires oxygen for aerobic respiration to grow and synthesize penicillin. Continually bubbling sterile air provides oxygen and assists in mixing the contents of the fermenter to keep the nutrients and fungi uniformly suspended.
评分标准
Award 1 mark for recognizing the dual role of bubbled sterile air in aerobic respiration and suspension mixing, option B.
题目 32 · 選擇題
1 分
Which hormone is responsible for maintaining the lining of the uterus during the second half of the menstrual cycle, and which organ secretes it?
A.FSH secreted by the pituitary gland
B.LH secreted by the ovary
C.oestrogen secreted by the pituitary gland
D.progesterone secreted by the corpus luteum
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解题
Progesterone is secreted by the corpus luteum (which forms in the ovary after ovulation). Its primary function during the second half of the menstrual cycle is to maintain and thicken the uterus lining in preparation for potential embryo implantation.
评分标准
Award 1 mark for identifying the hormone as progesterone and the secreting structure as the corpus luteum, option D.
题目 33 · 選擇題
1 分
Which statement explains why nerve impulses can only travel in one direction across a synapse?
A.Neurotransmitter molecules are active only in the synaptic cleft.
B.Postsynaptic neurones contain mitochondria to produce energy for active transport of neurotransmitters.
C.Receptors for neurotransmitters are only found on the postsynaptic membrane.
D.Vesicles containing neurotransmitters are found on both sides of the synaptic cleft.
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解题
Receptors for neurotransmitters are located exclusively on the postsynaptic membrane, while the neurotransmitter vesicles are only located in the presynaptic neurone. This ensures that the chemical signal can only be transmitted in one direction.
评分标准
1 mark for the correct option C.
题目 34 · 選擇題
1 分
An animal has a diploid chromosome number of 48. Which row correctly identifies the number of chromosomes in its gametes, and the type of cell division that produced them?
A.24 chromosomes, meiosis
B.24 chromosomes, mitosis
C.48 chromosomes, meiosis
D.48 chromosomes, mitosis
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解题
Gametes are haploid cells, meaning they contain half the number of chromosomes of a diploid body cell. Therefore, the gamete contains \(48 / 2 = 24\) chromosomes. Gametes are formed through meiosis, which is a reduction division.
评分标准
1 mark for the correct option A.
题目 35 · 選擇題
1 分
A photomicrograph of a plant cell has a magnification of \(\times 400\). The measured length of the cell in the photomicrograph is \(24\text{ mm}\). What is the actual length of the cell in micrometres (\(\mu\text{m}\))?
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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解题
First, convert the measured length from millimetres to micrometres: \(24\text{ mm} = 24,000\mu\text{m}\). Using the formula: \(\text{actual size} = \text{image size} / \text{magnification}\), we get \(\text{actual size} = 24,000\mu\text{m} / 400 = 60\mu\text{m}\).
评分标准
1 mark for the correct calculation and conversion to option C.
题目 36 · 選擇題
1 分
A student sets up a potometer to investigate the rate of transpiration of a leafy shoot. Which environmental changes would cause the bubble in the potometer to move the fastest?
A.decreased light intensity and increased humidity
B.decreased temperature and decreased wind speed
C.increased light intensity and decreased humidity
D.increased humidity and increased wind speed
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解题
Increasing light intensity increases stomatal opening, allowing more water vapour to escape. Decreasing humidity increases the concentration gradient of water vapour between the inside of the leaf and the surrounding air, accelerating evaporation and transpiration.
评分标准
1 mark for the correct option C.
题目 37 · 選擇題
1 分
Anaerobic respiration in yeast is utilized in various industrial processes. Which products are formed during this process, and how are they used in industry?
A.carbon dioxide (to make bread rise) and ethanol (for biofuel)
B.carbon dioxide (for biofuel) and lactic acid (to make yoghurt)
C.lactic acid (to preserve food) and ethanol (to make beer)
D.oxygen (to ferment fruit) and carbon dioxide (to carbonate drinks)
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解题
Anaerobic respiration in yeast (fermentation) produces carbon dioxide and ethanol. Carbon dioxide gas causes bread dough to expand and rise, while ethanol is utilized as a biofuel and in alcoholic beverages.
评分标准
1 mark for the correct option A.
题目 38 · 選擇題
1 分
Four identical pieces of potato tissue are placed in sucrose solutions of different concentrations: \(0.0\text{ mol/dm}^3\), \(0.2\text{ mol/dm}^3\), \(0.4\text{ mol/dm}^3\), and \(0.8\text{ mol/dm}^3\). In which concentration of sucrose solution would the potato cells become the most plasmolysed?
A.\(0.0\text{ mol/dm}^3\)
B.\(0.2\text{ mol/dm}^3\)
C.\(0.4\text{ mol/dm}^3\)
D.\(0.8\text{ mol/dm}^3\)
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解题
Plasmolysis occurs when plant cells lose water via osmosis. The highest concentration of sucrose solution (\(0.8\text{ mol/dm}^3\)) has the lowest water potential. Water leaves the potato cells down a water potential gradient, causing the cell membrane to pull away from the cell wall.
评分标准
1 mark for the correct option D.
题目 39 · 選擇題
1 分
During the cardiac cycle, the heart chambers contract and relax to pump blood. What is the state of the heart valves when the ventricles contract?
A.Atrioventricular valves are closed and semilunar valves are closed.
B.Atrioventricular valves are closed and semilunar valves are open.
C.Atrioventricular valves are open and semilunar valves are closed.
D.Atrioventricular valves are open and semilunar valves are open.
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解题
When the ventricles contract, high pressure forces the atrioventricular valves to close to prevent backflow of blood into the atria. At the same time, this high pressure forces the semilunar valves to open, allowing blood to exit the heart into the aorta and pulmonary artery.
评分标准
1 mark for the correct option B.
题目 40 · 選擇題
1 分
In a certain plant species, flower color is controlled by a gene with codominant alleles. Homozygous plants with the allele \(C^R\) have red flowers, homozygous plants with the allele \(C^W\) have white flowers, and heterozygous plants (\(C^R C^W\)) have pink flowers. If two pink-flowered plants are crossed, what is the expected ratio of flower colors in the offspring?
A.all pink
B.1 red : 1 white
C.1 red : 2 pink : 1 white
D.3 red : 1 white
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解题
Crossing two pink-flowered heterozygous plants (\(C^R C^W \times C^R C^W\)) yields the genotypic ratio \(1\ C^R C^R : 2\ C^R C^W : 1\ C^W C^W\). Since the alleles are codominant, this translates directly to a phenotypic ratio of 1 red : 2 pink : 1 white flower color.
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 题目 · 80 分
题目 1 · Structured
13 分
Figure 1.1 shows a diagram of a synapse between two neurones.
(a) Define the term *synapse*. [2]
(b) Describe the sequence of events that occurs when an electrical impulse reaches the synaptic knob of the presynaptic neurone. [6]
(c) Explain how the structure of a synapse ensures that nerve impulses travel in one direction only. [3]
(d) Some drugs can affect transmission across a synapse. State two ways in which a drug can reduce transmission across a synapse. [2]
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解题
(a) A synapse is a junction or junctional gap between two neurones, across which signals are transmitted via chemical messengers (neurotransmitters).
(b) When an electrical impulse reaches the presynaptic knob: 1. It stimulates the synaptic vesicles to fuse with the presynaptic membrane. 2. Neurotransmitter molecules are released into the synaptic cleft by exocytosis. 3. The neurotransmitter molecules diffuse across the synaptic cleft. 4. They bind to specific, complementary receptor proteins on the post-synaptic membrane. 5. This binding triggers a new electrical impulse in the post-synaptic neurone. 6. The neurotransmitter is then broken down by enzymes or reabsorbed.
(c) The structure ensures one-way travel because: - Neurotransmitter vesicles are only present in the presynaptic neurone. - Receptor proteins are only located on the post-synaptic membrane. - Therefore, diffusion can only happen from the presynaptic side to the post-synaptic side, meaning an impulse cannot be initiated in the reverse direction.
(d) A drug can reduce synaptic transmission by: - Blocking the receptor proteins on the post-synaptic membrane so neurotransmitters cannot bind. - Inhibiting the release of neurotransmitter molecules from the presynaptic vesicles.
评分标准
(a) Max 2 marks: - Junction / gap between two neurones [1] - Allows transmission of signals / impulses via chemical messengers [1]
(b) Max 6 marks: - Impulse triggers release of neurotransmitter molecules [1] - From synaptic vesicles [1] - Release via fusion with presynaptic membrane / exocytosis [1] - Neurotransmitters diffuse across the synaptic cleft / gap [1] - Bind to specific / complementary receptors [1] - On the post-synaptic membrane [1] - Triggers a new electrical impulse in the post-synaptic neurone [1] - Neurotransmitter is broken down / reabsorbed [1]
(c) Max 3 marks: - Vesicles / neurotransmitters only present in the presynaptic neurone / terminal [1] - Receptors only present on the post-synaptic membrane [1] - Diffusion of chemical messengers can only occur down a concentration gradient from pre- to post-synaptic neurone [1]
(d) Max 2 marks (any two of): - Blocking receptor sites on post-synaptic membrane [1] - Preventing release of neurotransmitters from vesicles [1] - Increasing the rate of neurotransmitter breakdown in the cleft [1] - Preventing vesicle fusion with the presynaptic membrane [1]
题目 2 · Structured
13 分
Cylinders of sweet potato tissue were placed in sucrose solutions of different concentrations. The sweet potato cylinders had an initial length of 50 mm. After 2 hours, the change in length of each cylinder was measured. Table 2.1 shows the results.
(a) Use the equation: $$\text{Percentage change in length} = \frac{\text{Final length} - \text{Initial length}}{\text{Initial length}} \times 100$$ Calculate the percentage change in length for the sweet potato cylinder placed in the 0.8 mol per dm\(^{3}\) sucrose solution. Show your working. [2]
(b) Explain, in terms of water potential, why the cylinder in 0.0 mol per dm\(^{3}\) sucrose solution increased in length. [4]
(c) State the sucrose concentration at which there was no net movement of water into or out of the sweet potato cells. Explain your answer. [3]
(d) Describe the state of the cells of the sweet potato cylinder after being placed in the 0.8 mol per dm\(^{3}\) sucrose solution, and explain how this state arises. [4]
(b) The 0.0 mol per dm\(^{3}\) sucrose solution (distilled water) has a higher water potential than the sweet potato cells. Water moves into the sweet potato cells by osmosis, down a water potential gradient, across a partially permeable membrane. This increases the turgor pressure inside the cells, stretching the cell walls and increasing the overall length of the cylinder.
(c) At 0.4 mol per dm\(^{3}\), because there was no change in length (final length remained 50.0 mm). This indicates that the water potential of the sucrose solution was equal to the water potential inside the sweet potato cells, so no net osmosis occurred.
(d) The sweet potato cells become plasmolysed (or flaccid). The 0.8 mol per dm\(^{3}\) sucrose solution has a lower water potential than the cytoplasm of the sweet potato cells. Water moves out of the cells by osmosis down a water potential gradient. The vacuole and cytoplasm shrink, causing the cell membrane to pull away from the cell wall.
评分标准
(a) Max 2 marks: - Correct calculation shown: \((46.5 - 50.0) / 50.0 \times 100\) [1] - Correct final answer: -7.0% (accept 7% decrease, ignore lack of minus sign if 'decrease' is stated) [1]
(b) Max 4 marks: - Sucrose solution (0.0 mol per dm\(^{3}\)) has a higher water potential than inside sweet potato cells [1] - Water moves into the sweet potato cells [1] - By osmosis [1] - Across a partially permeable membrane [1] - Down a water potential gradient [1] - Increase in turgor pressure / cells become turgid, stretching cell walls [1]
(c) Max 3 marks: - 0.4 mol per dm\(^{3}\) [1] - No change in length of cylinder / length remained 50 mm [1] - Sucrose solution and cytoplasm are isotonic / have equal water potentials [1]
(d) Max 4 marks: - Cells become plasmolysed / flaccid [1] - Sucrose solution (0.8 mol per dm\(^{3}\)) has a lower water potential than the sweet potato cell cytoplasm [1] - Water moves out of the cells [1] - Down a water potential gradient, by osmosis [1] - Cell membrane pulls away from the cell wall [1]
题目 3 · Structured
13 分
Chlamydia is a common sexually transmitted infection (STI).
(a) State the name of the pathogen that causes chlamydia and identify the group of organisms it belongs to. [2]
(b) Describe how chlamydia is transmitted between humans. [2]
(c) Explain the consequences of an untreated chlamydia infection on the human reproductive system. [4]
(d) Describe and explain three methods used to control the spread of STIs such as chlamydia. [5]
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解题
(a) Chlamydia is caused by the bacterium *Chlamydia trachomatis*, which belongs to the bacteria (prokaryote) kingdom.
(b) It is transmitted primarily through unprotected sexual intercourse (vaginal, anal, or oral sex) or from an infected mother to her baby during childbirth.
(c) Untreated chlamydia can spread to the reproductive organs, causing Pelvic Inflammatory Disease (PID) in females. This can result in inflammation and scarring of the oviducts (Fallopian tubes), leading to ectopic pregnancies or complete infertility in both males and females.
(d) Control methods: 1. **Use of barrier contraception (condoms):** Physically prevents the exchange of bodily fluids during sexual contact. 2. **Screening and diagnostic testing:** Identifies asymptomatic individuals so they can be treated and stop transmitting the pathogen. 3. **Contact tracing and antibiotic treatment:** Ensures partners of infected individuals are identified and cured, breaking the transmission chain.
(b) Max 2 marks: - Unprotected sexual intercourse / exchange of bodily fluids [1] - From mother to baby during vaginal delivery [1]
(c) Max 4 marks: - Can cause pelvic inflammatory disease (PID) in women [1] - Causes inflammation / scarring of the oviducts / Fallopian tubes [1] - Leads to blocked oviducts preventing fertilisation / infertility [1] - Increases risk of ectopic pregnancy [1] - Can cause pain / inflammation in testicles (epididymitis) in men [1]
(d) Max 5 marks: - Use of condoms / barrier methods [1]; physically blocks transmission of pathogens / bodily fluids [1] - Education / raising awareness [1]; increases safe sex practices [1] - Routine screening / testing [1]; detects asymptomatic carriers so they can be treated [1] - Contact tracing [1]; identifies and treats sexual partners of infected patients [1] - Treatment with antibiotics [1]; kills the pathogen, curing the individual and preventing further spread [1]
题目 4 · Structured
13 分
Hemophilia is a sex-linked genetic disorder caused by a recessive allele (\(X^{h}\)) on the X chromosome. The dominant allele for normal blood clotting is represented by \(X^{H}\). The Y chromosome does not carry an allele for this gene.
(a) Define the term *sex-linked characteristic*. [2]
(b) A woman who is a carrier for hemophilia has a child with a man who has normal blood clotting.
(i) State the genotypes of both parents. [2]
(ii) Draw a genetic diagram to determine the probability that their children will have hemophilia. Include: - Parental genotypes - Gametes - Offspring genotypes - Offspring phenotypes [5]
(iii) State the probability that a male child from these parents will have hemophilia. [1]
(c) Explain why males are more likely to suffer from sex-linked genetic disorders like hemophilia than females. [3]
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解题
(a) A sex-linked characteristic is a phenotype or trait determined by a gene located on one of the sex chromosomes (usually the X chromosome), which makes it exhibit different inheritance patterns in males and females.
(b)(ii) - **Parental genotypes:** \(X^{H}X^{h} \times X^{H}Y\) - **Gametes:** Mother: \(X^{H}\) and \(X^{h}\); Father: \(X^{H}\) and \(Y\) - **Offspring Genotypes:** \(X^{H}X^{H}\), \(X^{H}X^{h}\), \(X^{H}Y\), \(X^{h}Y\) - **Offspring Phenotypes:** - \(X^{H}X^{H}\): Normal female - \(X^{H}X^{h}\): Carrier female (normal clotting) - \(X^{H}Y\): Normal male - \(X^{h}Y\): Hemophiliac male - **Probability of hemophilia:** 25% (or 0.25 or 1 in 4).
(b)(iii) Among male children (\(X^{H}Y\) and \(X^{h}Y\)), the probability of having hemophilia is 50% (or 0.5).
(c) Males have only one X chromosome (XY). If they inherit the recessive allele (\(X^{h}\)) from their mother, they will suffer from the condition because they do not have a second X chromosome that could carry a dominant healthy allele (\(X^{H}\)) to mask it. Females have two X chromosomes (XX) and must inherit two recessive alleles to have the disorder.
评分标准
(a) Max 2 marks: - Gene responsible is located on a sex chromosome / X chromosome [1] - Trait is more common in one sex than the other [1]
(b)(ii) Max 5 marks: - Correct gametes shown: Mother: \(X^{H}\) and \(X^{h}\); Father: \(X^{H}\) and \(Y\) [1] - Correct offspring genotypes: \(X^{H}X^{H}\), \(X^{H}X^{h}\), \(X^{H}Y\), \(X^{h}Y\) [1] - Correct matching of phenotypes to genotypes: \(X^{H}X^{H}\) (normal female), \(X^{H}X^{h}\) (carrier/normal female), \(X^{H}Y\) (normal male), \(X^{h}Y\) (hemophiliac male) [1] - Indication of probability of hemophiliac offspring: 25% / 0.25 / 1 in 4 [1] - Genetic diagram layout is clear and structured [1]
(b)(iii) Max 1 mark: - 50% / 0.5 / 1 in 2 [1]
(c) Max 3 marks: - Males have only one X chromosome / are XY [1] - If they inherit the affected allele, they will exhibit the disease as there is no second allele to mask it [1] - Females have two X chromosomes / are XX and require two copies of the recessive allele to show the disorder [1]
题目 5 · Structured
14 分
The fungus *Penicillium* is grown in industrial fermenters to produce penicillin, an antibiotic.
(a) Name the type of organism that *Penicillium* is. [1]
(b) Explain why the following components of a fermenter are necessary for the successful growth of *Penicillium*:
(i) Water jacket [2]
(ii) Air filter [2]
(iii) Stirring paddles [2]
(c) Penicillin is a secondary metabolite. It is produced mainly during the stationary phase of the fungus's growth.
(i) Describe the characteristics of the stationary phase in a population growth curve. [3]
(ii) Explain why the population of *Penicillium* enters a stationary phase. [2]
(d) Suggest why penicillin is only produced when nutrients in the fermenter begin to run out. [2]
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解题
(a) *Penicillium* is a fungus.
(b)(i) The water jacket circulates cold water to absorb metabolic heat produced by the respiring fungus. This maintains a constant optimum temperature and prevents enzymes from denaturing.
(b)(ii) The air filter sterilises the incoming air, removing wild bacteria and spores. This prevents contamination and competition inside the fermenter vessel.
(b)(iii) Stirring paddles mix the culture medium to ensure even distribution of nutrients, oxygen, and temperature, and to prevent the fungal mycelium from settling at the bottom.
(c)(i) During the stationary phase, the birth/reproduction rate of the fungus equals its death rate. The overall population size remains constant and reaches a plateau because resources have become limiting.
(c)(ii) The population enters this phase because nutrients (like glucose and nitrogen) are running out, and toxic waste metabolites are accumulating in the fermenter.
(d) Penicillin is a secondary metabolite. It is produced under stress (nutrient depletion) to inhibit the growth of potential bacterial competitors in the environment, helping the fungus secure remaining resources.
评分标准
(a) Max 1 mark: - Fungus / fungi [1]
(b)(i) Max 2 marks: - To maintain optimum temperature / prevent overheating [1] - Heat from respiration can denature enzymes [1]
(b)(ii) Max 2 marks: - To prevent contamination / maintain a pure culture [1] - Avoids competition for nutrients / introduction of pathogens [1]
(b)(iii) Max 2 marks: - To distribute nutrients / oxygen / heat evenly throughout [1] - Prevents settling / clumping of the fungus [1]
(c)(i) Max 3 marks: - Rate of reproduction equals death rate [1] - Population size remains constant / plateaus [1] - Resources are limiting [1]
(c)(ii) Max 2 marks (any two of): - Depletion of nutrients (such as glucose / nitrogen) [1] - Accumulation of toxic metabolic wastes [1] - Lack of space / physical space limit [1]
(d) Max 2 marks: - Penicillin is a secondary metabolite / not needed for basic growth [1] - Produced as a defense mechanism to kill bacterial competitors when resources become scarce [1]
题目 6 · Structured
14 分
Cholera is an acute diarrheal disease caused by infection of the intestine.
(a) Describe how the cholera bacterium causes diarrhea and dehydration. [4]
(b) Explain why boiling drinking water is an effective method to prevent the spread of cholera. [2]
(c) Vaccination can protect individuals from cholera by stimulating active immunity.
(i) Define the term *active immunity*. [2]
(ii) Explain how a vaccine results in long-term active immunity against cholera. [4]
(d) In some situations, passive immunity is used to protect individuals. Distinguish between active immunity and passive immunity. [2]
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解题
(a) The cholera bacterium multiplies in the small intestine and releases a cholera toxin. This toxin stimulates the cells of the intestinal wall to actively pump chloride ions into the lumen of the intestine. The build-up of chloride ions lowers the water potential in the gut lumen. Water moves out of the blood and intestinal cells by osmosis down a water potential gradient, causing watery stools (diarrhea) and severe dehydration.
(b) Boiling water uses high temperatures to denature the critical proteins and enzymes of the cholera bacteria, killing them and rendering the water pathogen-free and safe to drink.
(c)(i) Active immunity is the defense against a pathogen by antibody production in the body, which can be acquired naturally through infection or artificially through vaccination.
(c)(ii) The vaccine contains weakened, killed, or fragments of the cholera pathogen containing antigens. These antigens trigger an immune response where lymphocytes produce complementary antibodies. Some of these lymphocytes develop into memory cells. If the person is exposed to the live pathogen later, these memory cells quickly recognise the antigens and produce massive amounts of antibodies rapidly, preventing disease.
(d) Active immunity involves the body making its own antibodies and memory cells, providing long-term protection. Passive immunity involves receiving pre-formed antibodies from another source (e.g., breast milk), providing immediate but short-term protection without memory cells.
评分标准
(a) Max 4 marks: - Bacteria colonise the small intestine and produce a toxin [1] - Toxin causes secretion of chloride ions into the intestinal lumen [1] - Lowers water potential inside the gut lumen [1] - Water moves out of blood / cells into the lumen [1] - By osmosis, down a water potential gradient [1] - Results in watery diarrhoea / dehydration [1]
(b) Max 2 marks: - High temperature kills the cholera bacteria [1] - Denatures proteins / enzymes of the pathogen [1]
(c)(i) Max 2 marks: - Defence against a pathogen by antibody production in the body [1] - Gained after infection / vaccination [1]
(c)(ii) Max 4 marks: - Vaccine contains dead / weakened / harmless cholera pathogens or antigens [1] - Stimulates lymphocytes to produce specific / complementary antibodies [1] - Mitotic division of lymphocytes produces clones [1] - Memory cells are formed [1] - Rapid and large-scale antibody production occurs upon future exposure [1]
(d) Max 2 marks: - Active immunity involves the body producing its own antibodies and memory cells, whereas passive immunity receives antibodies from outside [1] - Active immunity provides long-term protection, whereas passive immunity is short-lived [1]
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 题目 · 39.99 分
题目 1 · Experimental and Practical Analysis
13.33 分
A student investigated the leakage of red pigment (betalain) from beetroot cell vacuoles when exposed to different concentrations of ethanol.
The cell membrane is a selectively permeable structure that can be damaged by organic solvents like ethanol, causing intracellular pigments to leak out into the surrounding solution.
The student used the following method: - Beetroot cylinders were cut using a cork borer and sliced into thin discs of equal thickness. - The discs were rinsed in distilled water and dried with a paper towel. - Five test-tubes were prepared, each containing 10 cm³ of different concentrations of ethanol: 0%, 10%, 20%, 30%, and 40%. - Five beetroot discs were placed in each test-tube and left for 20 minutes. - After 20 minutes, the tubes were shaken, and the beetroot discs were removed. - The color intensity of the resulting liquid was compared to a standard color chart with values from 1 (completely clear/no leakage) to 10 (dark intense red/maximum leakage).
The results obtained were: - For 0% ethanol, the color intensity score was 1. - For 10% ethanol, the color intensity score was 3. - For 20% ethanol, the color intensity score was 5. - For 30% ethanol, the color intensity score was 8. - For 40% ethanol, the color intensity score was 10.
(a) Identify the independent variable and the dependent variable in this investigation. [2]
(b) Prepare a table to record these results. [4]
(c) State a conclusion for these results. [1]
(d) Explain why the beetroot discs were thoroughly rinsed with distilled water before starting the experiment. [2]
(e) Describe how you would test the liquid in the 40% ethanol tube at the end of the experiment for the presence of reducing sugars. [3]
(f) State one safety precaution required when working with ethanol. [1]
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解题
(a) Independent variable: concentration of ethanol (%). Dependent variable: color intensity score (from 1 to 10).
(b) Table containing two columns with suitable headings, including units where appropriate, showing all 5 concentrations and their corresponding color intensity scores.
(c) As the concentration of ethanol increases, the color intensity of the solution increases / more red pigment leaks out of the beetroot cells.
(d) Cutting the beetroot discs damages the cells at the cut edges, releasing pigment. Rinsing removes this leaked pigment so that any color change observed during the experiment is solely due to the effect of the ethanol.
(e) Add an equal volume of Benedict's reagent to the solution in a test-tube. Heat the mixture in a hot water-bath (above 80 °C) for 5 minutes. Observe the color change from blue to green, yellow, orange, or brick-red.
(f) Keep ethanol away from open flames / naked Bunsen burner flames because it is highly flammable (use a water-bath to heat instead).
评分标准
Total Marks: 13
(a) [2 marks] - Independent variable: concentration of ethanol (%); [1] - Dependent variable: color intensity (score / rating); [1]
(b) [4 marks] - Table drawn with neat lines and columns/rows clearly demarcated; [1] - Column headings: "Ethanol concentration / %" and "Color intensity (score 1-10)" (must have units for concentration); [1] - All 5 data points recorded correctly; [1] - Logically ordered by ascending concentration; [1]
(d) [2 marks] - To wash away pigment released from cells damaged during cutting; [1] - To ensure any pigment leaked during the experiment is only due to the treatment (membrane damage by ethanol); [1]
(e) [3 marks] - Add Benedict's reagent / solution; [1] - Heat / warm in a water-bath (above 70 °C); [1] - Correct color change observation (blue to green/yellow/orange/brick-red); [1]
(f) [1 mark] - Keep away from naked flames / use electric water-bath / wear safety goggles because ethanol is flammable; [1]
题目 2 · Experimental and Practical Analysis
13.33 分
Some crop plants are sensitive to soil salinity. A student wants to investigate the tolerance of mung bean seeds to sodium chloride (salt) solutions during germination.
(a) Plan an investigation to determine the effect of different concentrations of sodium chloride solution on the percentage germination of mung bean seeds. [6]
(b) A student carried out this investigation and obtained the following results for a sodium chloride solution of concentration 0.2 mol/dm³: - Petri Dish 1: 18 out of 20 seeds germinated. - Petri Dish 2: 17 out of 20 seeds germinated. - Petri Dish 3: 19 out of 20 seeds germinated.
Calculate the mean percentage germination of the mung bean seeds in this concentration of salt solution. Show your working. [3]
(c) Describe the role of the cotyledon during the early stages of seed germination before leaves have developed. [2]
(d) State two environmental conditions, other than water, required for seed germination. [2]
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解题
(a) Plan: - Use a range of at least 5 different concentrations of sodium chloride solution (e.g., 0.0, 0.2, 0.4, 0.6, 0.8 mol/dm³). - Place a fixed number of mung bean seeds (e.g., 20 seeds) on filter paper inside a Petri dish for each concentration. - Add a fixed volume of the respective salt solution to each Petri dish (e.g., 5 cm³). - Keep environmental factors constant: same temperature (e.g., 20 °C, incubator) and same light conditions. - Leave the dishes for a set period of time (e.g., 5 days). - Count the number of germinated seeds in each dish and calculate the percentage germination. - Repeat the experiment at least twice for each concentration to calculate a mean and identify anomalous results. - Safety precaution: wash hands after handling seeds / use sterile tweezers.
(c) The cotyledon contains stored food reserves (starch/proteins/lipids) which are broken down by enzymes (amylase/proteases) during germination to provide energy and raw materials for the growth of the radicle and plumule.
(d) Oxygen (for aerobic respiration) and a suitable temperature (for optimum enzyme activity).
评分标准
Total Marks: 13
(a) [6 marks] - independent variable: range of at least 5 salt concentrations (must specify values or say at least 5); [1] - dependent variable: count / record number of germinated seeds to find percentage; [1] - constant variable 1: volume of solution added / type of seed / number of seeds per dish; [1] - constant variable 2: temperature / light level / duration of experiment; [1] - replication: repeat at least 3 times (3 Petri dishes per concentration) to obtain mean; [1] - safety/hygiene: wash hands after handling soil/seeds / wear gloves / cut away on safe surface; [1]
(b) [3 marks] - MP1: Sum of germinated seeds (54) or individual percentages (90%, 85%, 95%); [1] - MP2: Correct formula dividing by total seeds (60) or dividing sum of percentages by 3; [1] - MP3: Final answer = 90%; [1]
(c) [2 marks] - Holds food stores / starch / proteins; [1] - Provides energy / nutrients / glucose for growth / respiration of the embryo (until photosynthesis starts); [1]
A student investigated the rate of transpiration in a leafy shoot over a 12-hour period.
(a) Fig. 3.1 shows a diagram of a stomatal pore on the lower surface of a leaf. A line AB is drawn across the stomatal pore.
Assume the measured length of line AB on the drawing is 45 mm.
Calculate the actual width of the stomatal pore if the magnification of the drawing is \(\times 1200\). Show your working and give your answer in micrometres (\(\mu m\)). [3]
(b) The average rate of transpiration of the leafy shoot was measured at different times of day. The results are shown in Table 3.1.
(b) Graph plotting requirements: - Axes labelled: x-axis "Time of day / hours" and y-axis "Transpiration rate / g/h". - Linear scales starting at sensible origins covering more than half the grid. - All points plotted accurately within half a small square. - Points joined with straight lines or a smooth curve.
(c) Trend description & explanation: - Rate of transpiration increases from 2.0 g/h to 23.0 g/h. - Because temperature increases / light intensity increases / humidity decreases. - Higher light intensity causes stomata to open wider / higher temperature increases kinetic energy of water molecules, increasing evaporation from mesophyll cells.
(d) Transpiration rate would decrease. An increase in relative humidity increases the water vapor concentration in the air outside the leaf. This decreases the water potential gradient / diffusion gradient between the air spaces inside the leaf and the external atmosphere, reducing the rate of diffusion of water vapor out of the stomata.
评分标准
Total Marks: 14
(a) [3 marks] - MP1: Correct conversion formula or actual size in mm (0.0375 mm); [1] - MP2: Multiplication by 1000 to convert to micrometres; [1] - MP3: Correct final answer = 37.5 (\(\mu m\)); [1]
(b) [4 marks] - Axes labelled with units on both axes (Time of day / hours AND Transpiration rate / g/h); [1] - Suitable linear scales occupying at least half the grid in both directions; [1] - All 7 points plotted accurately (± half a small square); [1] - Clean, thin line joining all points sequentially (no feathering, no extrapolation); [1]
(c) [3 marks] - Description: Transpiration rate increases (from 2.0 to 23.0 g/h / by 21.0 g/h); [1] - Explanation 1: Increase in temperature / light intensity (increases evaporation/stomata opening); [1] - Explanation 2: More kinetic energy of water vapor / steeper diffusion gradient; [1]
(d) [4 marks] - Rate of transpiration decreases; [1] - Higher humidity means more water vapor in the atmosphere; [1] - Decreases the water potential / concentration gradient between inside and outside leaf; [1] - Slower rate of diffusion of water vapor out of the stomata; [1]
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