Cambridge IGCSE · thinka 原创模拟试题

2023 Cambridge IGCSE Biology (0610) 模拟试题及答案详解

Thinka Nov 2023 (V1) Cambridge IGCSE-Style Mock — Biology (0610)

160 180 分钟2023
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V1) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

Paper 21 (選擇題)

Answer all forty multiple-choice questions. For each question there are four possible answers, A, B, C and D.
40 题目 · 40
题目 1 · 選擇題
1
An individual walks from a brightly lit outdoor garden into a dark corridor. Which row describes the changes that occur in the iris muscles and the diameter of the pupil?
  1. A.circular muscles contract, radial muscles relax, pupil diameter decreases
  2. B.circular muscles contract, radial muscles relax, pupil diameter increases
  3. C.circular muscles relax, radial muscles contract, pupil diameter decreases
  4. D.circular muscles relax, radial muscles contract, pupil diameter increases Jefferys pupil reflex in humans.
查看答案详解

解题

In dim light or darkness, the circular muscles of the iris relax and the radial muscles contract. This causes the pupil to dilate (pupil diameter increases) to allow more light to enter the eye.

评分标准

1 mark for the correct option. Correct answer: D.
题目 2 · 選擇題
1
A student accidentally touches a hot object and quickly pulls their hand away. What is the correct pathway of the nervous impulse in this reflex action?
  1. A.receptor -> motor neurone -> relay neurone -> sensory neurone -> effector
  2. B.receptor -> sensory neurone -> relay neurone -> motor neurone -> effector
  3. C.effector -> motor neurone -> relay neurone -> sensory neurone -> receptor
  4. D.effector -> sensory neurone -> relay neurone -> motor neurone -> receptor
查看答案详解

解题

The correct sequence of structures in a reflex arc is: receptor (detects stimulus) -> sensory neurone (transmits impulse to CNS) -> relay neurone (connects within CNS) -> motor neurone (transmits impulse to effector) -> effector (performs response).

评分标准

1 mark for the correct option. Correct answer: B.
题目 3 · 選擇題
1
Which list contains only substances that pass across the placenta from the maternal blood to the fetal blood?
  1. A.antibodies, glucose and oxygen
  2. B.antibodies, carbon dioxide and urea
  3. C.amino acids, carbon dioxide and glucose
  4. D.amino acids, oxygen and urea
查看答案详解

解题

Glucose, oxygen, amino acids, and antibodies move from maternal blood to fetal blood across the placenta. Waste products such as carbon dioxide and urea move in the opposite direction (from fetal blood to maternal blood).

评分标准

1 mark for the correct option. Correct answer: A.
题目 4 · 選擇題
1
During the human menstrual cycle, which hormone triggers ovulation and which gland secretes it?
  1. A.follicle-stimulating hormone (FSH) secreted by the ovary
  2. B.luteinising hormone (LH) secreted by the pituitary gland
  3. C.estrogen secreted by the pituitary gland
  4. D.progesterone secreted by the ovary
查看答案详解

解题

Ovulation is triggered by a surge of luteinising hormone (LH), which is produced and secreted by the pituitary gland.

评分标准

1 mark for the correct option. Correct answer: B.
题目 5 · 選擇題
1
Villi are specialized structures found in the small intestine that facilitate the absorption of digested food. Which row correctly matches a structural feature of a villus to its adaptation for absorption?
  1. A.Microvilli: increases surface area; Lacteal: absorbs fatty acids and glycerol; One-cell thick epithelium: shortens diffusion path
  2. B.Microvilli: secretes hydrochloric acid; Lacteal: absorbs glucose and amino acids; One-cell thick epithelium: protects from stomach acid
  3. C.Microvilli: increases surface area; Lacteal: absorbs glucose and amino acids; One-cell thick epithelium: secretes digestive enzymes
  4. D.Microvilli: secretes hydrochloric acid; Lacteal: absorbs fatty acids and glycerol; One-cell thick epithelium: shortens diffusion path
查看答案详解

解题

The microvilli on epithelial cells significantly increase the surface area available for absorption. The lacteal is a lymphatic capillary that absorbs fatty acids and glycerol. The single-cell-thick epithelium provides a very short distance for nutrients to diffuse across into the bloodstream.

评分标准

1 mark for the correct option. Correct answer: A.
题目 6 · 選擇題
1
Which conservation measure is most effective for managing wild fish stocks sustainably?
  1. A.decreasing the mesh size of commercial fishing nets
  2. B.introducing synthetic fertilisers into rivers to increase algae populations
  3. C.establishing designated marine protected areas where commercial fishing is banned
  4. D.increasing the quota for harvesting young, immature fish
查看答案详解

解题

Establishing marine protected areas (MPAs) allows fish populations to recover, feed, and reproduce without disturbance, helping to sustain overall stocks. Decreasing net mesh size is harmful as it catches immature fish before they can reproduce.

评分标准

1 mark for the correct option. Correct answer: C.
题目 7 · 選擇題
1
A population of yeast is grown in a laboratory flask with a constant supply of nutrients. After a period of rapid growth, the population reaches the stationary phase. What directly causes the population to enter this phase?
  1. A.The rate of cell division equals the rate of cell death.
  2. B.The rate of cell division exceeds the rate of cell death.
  3. C.The accumulation of toxic waste products completely stops cell death.
  4. D.The availability of space decreases the death rate to zero.
查看答案详解

解题

In the stationary phase of a growth curve, the rate of cell reproduction (division) equals the rate of cell death, resulting in a stable population size.

评分标准

1 mark for the correct option. Correct answer: A.
题目 8 · 選擇題
1
Which row correctly describes the homeostatic responses in a human skin when the external temperature drops?
  1. A.shunt vessels constrict, arterioles dilate, sweat secretion increases
  2. B.shunt vessels dilate, arterioles constrict, sweat secretion decreases
  3. C.shunt vessels constrict, arterioles constrict, sweat secretion increases
  4. D.shunt vessels dilate, arterioles dilate, sweat secretion decreases
查看答案详解

解题

When internal or external temperature drops, arterioles in the skin constrict (vasoconstriction) and shunt vessels dilate to reduce blood flow near the skin surface, conserving heat. Sweat secretion also decreases to minimise evaporative cooling.

评分标准

1 mark for the correct option. Correct answer: B.
题目 9 · 選擇題
1
When a person moves from a dark room into bright sunlight, their pupils constrict to protect the retina. Which row correctly describes the state of the iris muscles during this pupil reflex?
  1. A.circular muscles contract, radial muscles relax
  2. B.circular muscles relax, radial muscles contract
  3. C.circular muscles contract, radial muscles contract
  4. D.circular muscles relax, radial muscles relax
查看答案详解

解题

In bright light, the pupil constricts to reduce the amount of light entering the eye. This pupil reflex is controlled by antagonistic muscles in the iris. The circular muscles contract, while the radial muscles relax, making the pupil smaller.

评分标准

1 mark for correct option. Correct answer: A.
题目 10 · 選擇題
1
Which statement describes how an impulse is transmitted across a synapse from one neurone to the next?
  1. A.Electrical signals jump directly across the synaptic cleft.
  2. B.Neurotransmitter molecules diffuse across the synaptic cleft and bind to receptors.
  3. C.Sodium ions are actively transported across the gap through protein channels.
  4. D.Hormones are released from the myelin sheath to stimulate the next neurone.
查看答案详解

解题

A synapse is a junction between two neurones. When an electrical impulse reaches the end of the presynaptic neurone, it triggers the release of neurotransmitter chemicals. These molecules diffuse across the synaptic cleft (gap) and bind to specific receptor proteins on the postsynaptic membrane, generating a new electrical impulse.

评分标准

1 mark for correct option. Correct answer: B.
题目 11 · 選擇題
1
The placenta allows the exchange of substances between the maternal blood and the fetal blood. Which substance moves from the fetal blood to the maternal blood across the placenta?
  1. A.oxygen
  2. B.glucose
  3. C.carbon dioxide
  4. D.antibodies
查看答案详解

解题

Waste products from the fetus, such as carbon dioxide and urea, must be transported away. These diffuse from the fetal blood capillaries across the placenta into the maternal blood to be excreted by the mother. Beneficial substances like oxygen, glucose, and antibodies move in the opposite direction (from maternal to fetal blood).

评分标准

1 mark for correct option. Correct answer: C.
题目 12 · 選擇題
1
What is the primary effect of the sudden surge in luteinising hormone (LH) concentration during the female menstrual cycle?
  1. A.It stimulates the development of follicles in the ovary.
  2. B.It triggers the release of a mature egg cell from an ovary follicle.
  3. C.It causes the thickening and repair of the uterine lining.
  4. D.It inhibits the production of progesterone by the corpus luteum.
查看答案详解

解题

Luteinising hormone (LH) is produced by the pituitary gland. A sharp increase or surge in LH levels around day 14 of the cycle triggers ovulation, which is the release of a mature egg from its follicle in the ovary.

评分标准

1 mark for correct option. Correct answer: B.
题目 13 · 選擇題
1
Which structure in a villus is responsible for absorbing fatty acids and glycerol into the lymphatic system?
  1. A.lacteal
  2. B.capillary network
  3. C.goblet cell
  4. D.epithelium
查看答案详解

解题

Villi are tiny, finger-like projections in the small intestine that increase surface area for absorption. Each villus contains a central lymphatic vessel called a lacteal, which absorbs the products of fat digestion (fatty acids and glycerol). Other nutrients like glucose and amino acids are absorbed directly into the blood capillary network.

评分标准

1 mark for correct option. Correct answer: A.
题目 14 · 選擇題
1
Four statements about the adaptations of the small intestine for absorption are listed.

1. It has millions of villi which increase the surface area.
2. The epithelial wall of each villus is only one cell thick.
3. Lacteals absorb amino acids directly into the bloodstream.
4. It is long, providing a large surface area and more time for absorption.

Which statements are correct?
  1. A.1, 2 and 3 only
  2. B.1, 2 and 4 only
  3. C.1, 3 and 4 only
  4. D.2, 3 and 4 only
查看答案详解

解题

Statements 1, 2, and 4 correctly describe adaptations that facilitate efficient absorption in the small intestine. Statement 3 is incorrect because lacteals absorb fatty acids and glycerol into the lymphatic system, not amino acids (which are absorbed into blood capillaries).

评分标准

1 mark for correct option. Correct answer: B.
题目 15 · 選擇題
1
Which practice would be most effective as a conservation measure to sustainably manage wild fish stocks?
  1. A.decreasing the mesh size of commercial fishing nets
  2. B.increasing catch quotas during the main breeding season
  3. C.restricting fishing activities by establishing no-catch marine reserves
  4. D.introducing non-native predatory species to the marine habitat
查看答案详解

解题

Sustainably managing fish stocks involves maintaining population sizes high enough to allow reproduction and recovery. Establishing marine reserves (no-catch zones) protects habitats and allows populations of target species to spawn and rebuild. Decreasing mesh size catches younger, immature fish; increasing quotas during breeding seasons disrupts reproduction; and introducing non-native species can destroy native ecosystems.

评分标准

1 mark for correct option. Correct answer: C.
题目 16 · 選擇題
1
Why is it important to conserve ecosystems and maintain genetic diversity in wild populations?
  1. A.to reduce the resistance of wild species to environmental changes
  2. B.to provide potential resources for medical research and crop improvement
  3. C.to encourage inbreeding within small populations
  4. D.to accelerate the natural rate of species extinction
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解题

Conserving ecosystems and wild populations maintains a large gene pool (genetic diversity). This diversity is vital because it provides wild genes that can be bred into crops for pest/disease resistance, and organisms that contain chemical compounds of potential use in developing new medicines. Minimising inbreeding and reducing extinction rates are goals of conservation, not the opposite.

评分标准

1 mark for correct option. Correct answer: B.
题目 17 · 選擇題
1
A person walks out of a dark room into bright sunlight.

Which row shows the correct response of the iris muscles and the change in pupil diameter?

| | circular muscles of iris | radial muscles of iris | pupil diameter |
| :--- | :--- | :--- | :--- |
| **A** | contract | relax | decreases |
| **B** | contract | relax | increases |
| **C** | relax | contract | decreases |
| **D** | relax | contract | increases |
  1. A.A
  2. B.B
  3. C.C
  4. D.D
查看答案详解

解题

When a person enters a brightly lit environment, the pupil reflex prevents too much light from damaging the retina. The circular muscles of the iris contract, while the radial muscles relax. This antagonistic action causes the pupil diameter to decrease (constriction).

评分标准

1 mark for identifying row A as the correct combination of muscle actions and pupil change.
题目 18 · 選擇題
1
Which substances move from the maternal blood to the fetal blood across the placenta?
  1. A.carbon dioxide, urea, and oxygen
  2. B.glucose, oxygen, and antibodies
  3. C.glucose, amino acids, and urea
  4. D.oxygen, carbon dioxide, and hormones
查看答案详解

解题

Useful nutrients and protective molecules like glucose, oxygen, and antibodies pass from the maternal blood across the placenta to the fetal blood. In contrast, waste products like carbon dioxide and urea pass in the opposite direction (from fetal to maternal blood).

评分标准

1 mark for selecting the option containing only substances that move from maternal to fetal blood.
题目 19 · 選擇題
1
What is the main function of the lacteal in a villus of the small intestine?
  1. A.absorption of glucose and water
  2. B.absorption of amino acids and mineral ions
  3. C.absorption of fatty acids and glycerol
  4. D.transport of oxygenated blood to the liver
查看答案详解

解题

Each villus in the small intestine contains a network of blood capillaries and a central lymphatic vessel called a lacteal. The blood capillaries absorb water-soluble molecules like glucose and amino acids, whereas the lacteal is responsible for absorbing fat-soluble digested products, specifically fatty acids and glycerol.

评分标准

1 mark for identifying the absorption of fatty acids and glycerol as the function of the lacteal.
题目 20 · 選擇題
1
Which measures can be used to prevent the overexploitation of fish stocks to ensure sustainable fishing?

1. introducing a closed season for fishing
2. increasing the mesh size of nets
3. using drift nets with small mesh size
4. introducing legal quotas on the weight of fish caught
  1. A.1, 2 and 3 only
  2. B.1, 2 and 4 only
  3. C.1, 3 and 4 only
  4. D.2, 3 and 4 only
查看答案详解

解题

Sustainable fishing practices include introducing a closed season (1) to allow fish to reproduce without disruption, increasing net mesh size (2) to let young and undersized fish escape and grow to breeding age, and introducing legal catch quotas (4) to prevent excessive harvesting. Using small-mesh drift nets (3) is indiscriminate and causes overfishing, which is unsustainable.

评分标准

1 mark for identifying statements 1, 2, and 4 as correct measures.
题目 21 · 選擇題
1
What happens to the shunt vessels and arterioles in the skin when a person enters a cold environment?

| | shunt vessels | skin arterioles | blood flow to skin capillaries |
| :--- | :--- | :--- | :--- |
| **A** | constrict | dilate | increases |
| **B** | constrict | constrict | decreases |
| **C** | dilate | constrict | decreases |
| **D** | dilate | dilate | increases |
  1. A.A
  2. B.B
  3. C.C
  4. D.D
查看答案详解

解题

In a cold environment, the body responds by conserving core heat through vasoconstriction. Arterioles supplying skin capillaries constrict, reducing blood flow to the surface. Simultaneously, shunt vessels dilate to allow blood to bypass the superficial capillaries, thereby minimizing heat loss by radiation.

评分标准

1 mark for selecting row C as the correct physiological response to a cold environment.
题目 22 · 選擇題
1
Which row correctly identifies the hormones that repair and maintain the lining of the uterus during the menstrual cycle?

| | hormone that repairs the uterus lining | hormone that maintains the uterus lining |
| :--- | :--- | :--- |
| **A** | LH | FSH |
| **B** | oestrogen | progesterone |
| **C** | progesterone | oestrogen |
| **D** | FSH | LH |
  1. A.A
  2. B.B
  3. C.C
  4. D.D
查看答案详解

解题

During the follicular phase, oestrogen is secreted to repair and thicken the endometrium (uterus lining) after menstruation. Following ovulation, progesterone is secreted by the corpus luteum to maintain the thickened lining in preparation for potential fertilization and implantation.

评分标准

1 mark for identifying row B as the correct hormones for repair (oestrogen) and maintenance (progesterone) of the uterus lining.
题目 23 · 選擇題
1
Which row correctly matches a region of the digestive system with its function?

| | region of the digestive system | function |
| :--- | :--- | :--- |
| **A** | stomach | absorption of amino acids |
| **B** | gall bladder | production of bile |
| **C** | liver | production of bile |
| **D** | pancreas | absorption of water |
  1. A.A
  2. B.B
  3. C.C
  4. D.D
查看答案详解

解题

The liver is the organ responsible for producing bile, which is then stored in the gall bladder. The stomach primarily performs protein digestion and mechanical churning but not absorption of amino acids. The pancreas produces pancreatic juices (enzymes) but does not absorb water.

评分标准

1 mark for choosing row C as the correct matching pair.
题目 24 · 選擇題
1
Which environmental consequences can result directly from deforestation?

1. increased risk of flooding
2. increased concentration of carbon dioxide in the atmosphere
3. increased biodiversity
4. increased risk of soil erosion
  1. A.1, 2 and 3 only
  2. B.1, 2 and 4 only
  3. C.1, 3 and 4 only
  4. D.2, 3 and 4 only
查看答案详解

解题

Deforestation has several severe impacts on the environment: loss of roots leads to soil erosion (4); lack of tree canopy and transpiration increases run-off, which causes flooding (1); and removing photosynthesising trees causes carbon dioxide levels in the atmosphere to rise (2). Deforestation decreases biodiversity, so statement 3 is incorrect.

评分标准

1 mark for selecting B, indicating statements 1, 2, and 4 are correct consequences.
题目 25 · 選擇題
1
Which row shows the correct states of the iris muscles and the change in pupil size when a person walks from a dark room into a brightly lit corridor?
  1. A.circular muscles: contract; radial muscles: relax; pupil size: decreases
  2. B.circular muscles: relax; radial muscles: contract; pupil size: increases
  3. C.circular muscles: contract; radial muscles: contract; pupil size: decreases
  4. D.circular muscles: relax; radial muscles: relax; pupil size: increases block due to bright light stimulationed pupil reflex.
查看答案详解

解题

In bright light, the pupil constricts to reduce the amount of light entering the eye and protect the retina. This constriction is brought about by the contraction of the circular muscles and the relaxation of the radial muscles in the iris.

评分标准

Award 1 mark for the correct option A.
题目 26 · 選擇題
1
The table shows some features of synaptic transmission. Which row is correct?
  1. A.direction of movement: from presynaptic to postsynaptic neurone; location of receptors: postsynaptic membrane; mechanism of movement: diffusion
  2. B.direction of movement: from postsynaptic to presynaptic neurone; location of receptors: postsynaptic membrane; mechanism of movement: active transport
  3. C.direction of movement: from presynaptic to postsynaptic neurone; location of receptors: presynaptic membrane; mechanism of movement: osmosis
  4. D.direction of movement: from postsynaptic to presynaptic neurone; location of receptors: presynaptic membrane; mechanism of movement: diffusion
查看答案详解

解题

During synaptic transmission, neurotransmitters are released from the presynaptic neurone and diffuse across the synaptic cleft to bind to specific receptor proteins located on the postsynaptic membrane.

评分标准

Award 1 mark for the correct option A.
题目 27 · 選擇題
1
Which hormone is correctly paired with its main function during the human menstrual cycle?
  1. A.estrogen: stimulates the development of mature follicles
  2. B.FSH: maintains the thickness of the uterine lining
  3. C.LH: stimulates ovulation (the release of an egg)
  4. D.progesterone: repairs the uterine lining after menstruation
查看答案详解

解题

Luteinising hormone (LH) stimulates ovulation, which is the release of a mature egg cell from the follicle in the ovary. Follicle-stimulating hormone (FSH) stimulates follicle development, estrogen repairs and builds up the uterine lining, and progesterone maintains the thick uterine lining.

评分标准

Award 1 mark for the correct option C.
题目 28 · 選擇題
1
Which physical changes at puberty are stimulated by testosterone and estrogen?
  1. A.stimulated by testosterone only: growth of facial hair; stimulated by estrogen only: breast development; stimulated by both: growth of pubic hair
  2. B.stimulated by testosterone only: widening of the pelvis; stimulated by estrogen only: deepening of the voice; stimulated by both: growth of underarm hair
  3. C.stimulated by testosterone only: growth of pubic hair; stimulated by estrogen only: widening of the pelvis; stimulated by both: growth of facial hair
  4. D.stimulated by testosterone only: deepening of the voice; stimulated by estrogen only: growth of underarm hair; stimulated by both: breast development
查看答案详解

解题

At puberty, testosterone stimulates male-specific secondary sexual characteristics like facial hair growth. Estrogen stimulates female-specific characteristics like breast development. Both hormones stimulate the growth of pubic and underarm hair in their respective sexes.

评分标准

Award 1 mark for the correct option A.
题目 29 · 選擇題
1
Which row correctly identifies the main nutrients absorbed by the blood capillaries and the lacteals within a villus of the small intestine?
  1. A.absorbed by the blood capillary: amino acids and glucose; absorbed by the lacteal: fatty acids and glycerol
  2. B.absorbed by the blood capillary: fatty acids and glycerol; absorbed by the lacteal: amino acids and glucose
  3. C.absorbed by the blood capillary: glucose and glycogen; absorbed by the lacteal: fatty acids and amino acids
  4. D.absorbed by the blood capillary: starch and amino acids; absorbed by the lacteal: glucose and glycerol
查看答案详解

解题

The blood capillaries of the villus absorb water-soluble nutrients such as amino acids, glucose, minerals, and vitamins. The lacteals (lymph capillaries) absorb fat-soluble products, which are fatty acids and glycerol.

评分标准

Award 1 mark for the correct option A.
题目 30 · 選擇題
1
Which row correctly identifies the main organs where mechanical digestion of food and chemical digestion of lipids occur?
  1. A.main site(s) of mechanical digestion: mouth and stomach; main site of chemical digestion of lipids: small intestine
  2. B.main site(s) of mechanical digestion: mouth only; main site of chemical digestion of lipids: stomach
  3. C.main site(s) of mechanical digestion: stomach and small intestine; main site of chemical digestion of lipids: mouth
  4. D.main site(s) of mechanical digestion: small intestine only; main site of chemical digestion of lipids: large intestine
查看答案详解

解题

Mechanical digestion (physical breakdown of food) occurs mainly in the mouth (chewing) and stomach (churning). Chemical digestion of lipids occurs primarily in the small intestine, where lipase enzymes break down fats and oils into fatty acids and glycerol, aided by bile.

评分标准

Award 1 mark for the correct option A.
题目 31 · 選擇題
1
Some human actions in conservation are listed.
1. Using fishing nets with a large mesh size
2. Introducing a closed season for fishing during breeding periods
3. Increasing the use of chemical fertilisers in nearby agricultural fields

Which of these actions help to manage fish stocks sustainably?
  1. A.1 and 2 only
  2. B.1 and 3 only
  3. C.2 and 3 only
  4. D.1, 2 and 3
查看答案详解

解题

Nets with a large mesh size allow small, immature fish to escape and reach reproductive age. Closed seasons protect adult fish during their spawning periods. Chemical fertilisers run off into waterways and cause eutrophication, which decreases oxygen levels and kills fish rather than conserving them.

评分标准

Award 1 mark for the correct option A.
题目 32 · 選擇題
1
Which factor causes a population of yeast cells in a closed culture to enter the stationary phase of growth?
  1. A.depletion of nutrients and accumulation of toxic waste products
  2. B.a sudden increase in the concentration of dissolved oxygen
  3. C.decrease in the rate of cell death compared to the rate of cell division
  4. D.disappearance of all intraspecific competition
查看答案详解

解题

In a closed culture with limited resources, a yeast population eventually enters the stationary phase when cell division rate equals cell death rate. This is caused by the depletion of essential nutrients (like glucose) and the accumulation of toxic metabolic waste products (such as ethanol).

评分标准

Award 1 mark for the correct option A.
题目 33 · 選擇題
1
A student accidentally steps on a sharp pin and quickly lifts their foot. Which sequence shows the correct pathway of the nervous impulse during this reflex action?
  1. A.receptor \(\rightarrow\) sensory neurone \(\rightarrow\) relay neurone \(\rightarrow\) motor neurone \(\rightarrow\) effector
  2. B.effector \(\rightarrow\) motor neurone \(\rightarrow\) relay neurone \(\rightarrow\) sensory neurone \(\rightarrow\) receptor
  3. C.receptor \(\rightarrow\) motor neurone \(\rightarrow\) relay neurone \(\rightarrow\) sensory neurone \(\rightarrow\) effector
  4. D.effector \(\rightarrow\) sensory neurone \(\rightarrow\) relay neurone \(\rightarrow\) motor neurone \(\rightarrow\) receptor
查看答案详解

解题

During a reflex action, the stimulus is detected by a receptor, which generates an electrical impulse. This impulse travels along a sensory neurone to the central nervous system (spinal cord), where it is passed across a synapse to a relay neurone. From the relay neurone, the impulse is transmitted to a motor neurone, which carries it to an effector (such as a muscle) to produce a response. Therefore, the correct pathway is: receptor → sensory neurone → relay neurone → motor neurone → effector.

评分标准

Award 1 mark for the correct answer A.
题目 34 · 選擇題
1
When a person walks from a dark room into bright sunlight, which changes occur in the iris of the eye to protect the retina from damage?
  1. A.circular muscles contract, radial muscles relax, pupil size decreases
  2. B.circular muscles contract, radial muscles relax, pupil size increases
  3. C.circular muscles relax, radial muscles contract, pupil size decreases
  4. D.circular muscles relax, radial muscles contract, pupil size increases
查看答案详解

解题

In bright light, the pupil constricts (becomes smaller) to reduce the amount of light entering the eye. This pupil reflex is controlled by antagonistic muscles in the iris: the circular muscles contract and the radial muscles relax.

评分标准

Award 1 mark for the correct answer A.
题目 35 · 選擇題
1
Which group of substances passes from the mother's blood to the fetal blood across the placenta?
  1. A.carbon dioxide, urea and glucose
  2. B.carbon dioxide, oxygen and amino acids
  3. C.glucose, oxygen and antibodies
  4. D.urea, oxygen and antibodies
查看答案详解

解题

Useful substances like glucose, oxygen, amino acids, and maternal antibodies cross from the mother's blood to the fetus's blood via diffusion across the placenta. Waste products like carbon dioxide and urea move in the opposite direction (from fetus to mother).

评分标准

Award 1 mark for the correct answer C.
题目 36 · 選擇題
1
Which row correctly pairs a menstrual cycle hormone with its site of production and its main function?
  1. A.Progesterone | ovaries (corpus luteum) | maintains the uterus lining
  2. B.FSH | ovaries | stimulates ovulation
  3. C.LH | pituitary gland | stimulates the repair of the uterus lining
  4. D.Estrogen | pituitary gland | stimulates the development of follicles
查看答案详解

解题

Progesterone is produced by the corpus luteum in the ovaries and is responsible for maintaining the thickness and vascularisation of the uterus lining in preparation for a potential embryo.

评分标准

Award 1 mark for the correct answer A.
题目 37 · 選擇題
1
Villi are small projections in the small intestine that facilitate absorption. Which products of digestion are absorbed directly into the lacteal of a villus?
  1. A.glucose and amino acids
  2. B.fatty acids and glycerol
  3. C.glycogen and water-soluble vitamins
  4. D.starch and proteins
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解题

Lacteals are lymphatic capillaries situated in the center of intestinal villi. They specifically absorb fats, which are broken down into fatty acids and glycerol. Soluble nutrients like glucose, amino acids, minerals, and water-soluble vitamins are absorbed into the blood capillaries.

评分标准

Award 1 mark for the correct answer B.
题目 38 · 選擇題
1
Which features of the small intestine wall adapt it for a high rate of absorption of digested nutrients?

1. Villi and microvilli to increase surface area.
2. An epithelium that is only one cell thick to reduce diffusion distance.
3. A rich blood capillary network to maintain a steep concentration gradient.
  1. A.1, 2 and 3
  2. B.1 and 2 only
  3. C.1 and 3 only
  4. D.2 and 3 only
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解题

All three features are essential structural adaptations of the small intestine. Villi and microvilli provide a massive surface area for absorption. The single-cell-thick epithelium ensures a short distance for diffusion and active transport. The rich capillary network rapidly carries away absorbed nutrients, maintaining a steep concentration gradient between the lumen of the gut and the blood.

评分标准

Award 1 mark for the correct answer A.
题目 39 · 選擇題
1
Which conservation measure helps to maintain sustainable wild fish stocks specifically by preventing the capture of young, immature fish?
  1. A.introducing quota systems for total catch
  2. B.establishing closed seasons during breeding periods
  3. C.creating marine protected areas
  4. D.using nets with larger mesh sizes
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解题

Using nets with larger mesh sizes allows smaller, younger fish to escape through the gaps in the net. This ensures they survive to reach reproductive maturity and maintain the population. While quotas, closed seasons, and marine protected areas are also valid conservation tools, they do not specifically target the exclusion of small, immature fish from catches.

评分标准

Award 1 mark for the correct answer D.
题目 40 · 選擇題
1
Which of the following describes a key role of tropical rainforests in maintaining global climate stability?
  1. A.they prevent soil erosion by increasing surface runoff
  2. B.they release large amounts of methane during photosynthesis
  3. C.they act as major carbon sinks by absorbing carbon dioxide
  4. D.they decrease the rate of transpiration to reduce global rainfall
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解题

Tropical rainforests act as major carbon sinks because trees absorb large amounts of carbon dioxide from the atmosphere during photosynthesis, converting it into organic compounds. This helps regulate the greenhouse effect and stabilize global temperatures. They increase transpiration (not decrease it) to sustain local rainfall, and they reduce soil erosion by slowing down surface runoff.

评分标准

Award 1 mark for the correct answer C.

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Paper 41 (Theory Extended)

Answer all structured questions in the spaces provided. Show your working in calculations.
6 题目 · 80
题目 1 · structured
14
1 (a) The human eye adjusts to focus on near and distant objects.

(i) Name the process of focusing on objects at different distances.

(ii) Describe the changes that occur in the eye when focusing on a near object.

(b) Photoreceptors in the retina detect light intensity and colour.

(i) Compare the roles of rod cells and cone cells in vision.

(ii) Describe how an impulse is transmitted across a synapse from a photoreceptor cell to a sensory neurone.

(c) Some reflexes do not involve the brain.

(i) Define the term reflex action.

(ii) State one advantage of reflex actions.
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解题

(a) (i) Accommodation.
(ii) Ciliary muscles contract, suspensory ligaments slacken/become loose, the lens becomes more spherical/convex/thicker, and refracts light more strongly.

(b) (i) Rod cells are sensitive to low light intensity (night/dim vision) and do not detect colour (monochromatic/black and white). Cone cells require high light intensity (day/bright vision) and detect colour (three types: red, green, blue).
(ii) An electrical impulse arrives at the presynaptic membrane, stimulating vesicles to fuse with the membrane and release neurotransmitters into the synaptic cleft. The neurotransmitters diffuse across the cleft and bind to specific receptors on the postsynaptic membrane of the sensory neurone, generating a new electrical impulse.

(c) (i) A fast, automatic/involuntary, and protective response to a specific stimulus.
(ii) Rapid protection from damage/harm (survival).

评分标准

(a) (i)
1. Accommodation; [1]

(ii) (max 3 marks from:)
1. Ciliary muscles contract;
2. Suspensory ligaments become slack/loose;
3. Lens becomes more rounded/convex/thick/curved;
4. Light is refracted more strongly;
[3]

(b) (i) (max 3 marks from:)
1. Rods work in low light intensity/dim light AND cones work in high light intensity/bright light;
2. Rods give monochromatic/black and white vision AND cones detect colour/red, green, and blue light;
3. Rods are distributed across the retina/periphery AND cones are concentrated in the fovea;
[3]

(b) (ii) (max 4 marks from:)
1. Impulse/action potential reaches pre-synaptic membrane/knob;
2. Vesicles release neurotransmitters (into the cleft);
3. Neurotransmitters diffuse across the synaptic cleft;
4. Neurotransmitters bind to specific receptors on post-synaptic membrane;
5. Stimulates/triggers an impulse in the sensory neurone;
[4]

(c) (i)
1. Rapid/fast/immediate;
2. Involuntary/automatic/not requiring conscious thought;
[2]

(c) (ii)
1. Prevents injury/increases chances of survival/protects the body;
[1]
题目 2 · structured
13
2 (a) During the menstrual cycle, hormones regulate the thickening and shedding of the uterus lining.

(i) Name the hormone that stimulates the development of follicles in the ovary.

(ii) Explain the roles of progesterone and oestrogen in preparing the uterus for pregnancy.

(b) In-vitro fertilisation (IVF) is a fertility treatment used when couples have difficulty conceiving naturally.

(i) Outline the main steps involved in the process of IVF.

(ii) Discuss one social or ethical implication of IVF treatment.

(c) Sperm cells and egg cells are adapted for their functions.

State two structural differences between a sperm cell and an egg cell.
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解题

(a) (i) Follicle-stimulating hormone (FSH).
(ii) Oestrogen repairs and thickens the uterus lining/endometrium after menstruation. Progesterone maintains the thickened lining of the uterus, keeping it vascularized and ready for implantation of an embryo.

(b) (i) The woman is given FSH to stimulate multiple follicle development. Mature eggs are harvested from the ovaries. Sperm is collected from the man. Eggs and sperm are mixed in a Petri dish for fertilisation. Embryos are grown in a laboratory before one or two are transferred back into the woman's uterus.
(ii) Ethical issues include the creation and subsequent destruction of unwanted/surplus embryos, or the selective implantation. Social issues include financial cost, emotional stress, or enabling older/single parents to have children.

(c) Sperm cells have a flagellum (tail), are much smaller, contain an acrosome, and have very little cytoplasm compared to egg cells, which are large, immobile, contain a jelly coat (zona pellucida), and have a large food store (cytoplasm).

评分标准

(a) (i)
1. FSH / follicle-stimulating hormone; [1]

(ii) (max 4 marks from:)
1. Oestrogen repairs / thickens uterus lining;
2. Oestrogen stimulates LH secretion / inhibits FSH;
3. Progesterone maintains the uterus lining;
4. Progesterone prevents breakdown/shedding of the lining;
5. Prepared for implantation of the fertilised egg/embryo;
[4]

(b) (i) (max 4 marks from:)
1. Stimulation of ovaries/follicles using hormones (FSH);
2. Collection/harvesting of eggs/oocytes (from ovaries);
3. Fertilisation outside the body/in laboratory glass dish;
4. Incubation of embryos/growth of zygotes;
5. Transfer of embryos into the uterus/lining;
[4]

(b) (ii) (max 2 marks from:)
1. Destruction of surplus/unused embryos (ethical issue);
2. High cost/unequal access to treatment (social issue);
3. Emotional stress on the couple;
4. Use of donor gametes leading to parental identity issues;
[2]

(c)
1. Sperm has flagellum/tail OR egg is immobile;
2. Sperm is smaller OR egg is larger;
3. Sperm has acrosome OR egg has jelly coat;
4. Sperm has minimal cytoplasm OR egg has large nutrient-rich cytoplasm;
(Accept any two correct comparative points) [2]
题目 3 · structured
13
3 (a) The ileum is adapted for the absorption of digested food molecules.

(i) Explain how the structure of a villus increases the efficiency of absorption.

(ii) Describe the absorption of fatty acids and glycerol into the lymphatic system.

(b) Glucose and amino acids are absorbed into the capillary network of the villi.

(i) Distinguish between the mechanisms of diffusion and active transport in the absorption of glucose.

(ii) State the name of the blood vessel that transports these absorbed nutrients directly to the liver.

(c) Assimilation is the movement of digested food molecules into the cells of the body where they are used, becoming part of the cells.

State three ways the liver processes glucose after absorption.
查看答案详解

解题

(a) (i) Large surface area provided by microvilli/villi. Thin epithelium (one cell thick) which reduces the diffusion distance. Good blood capillary network to maintain a steep concentration gradient. Lacteal present in the centre to transport fats.
(ii) Fatty acids and glycerol diffuse into the epithelial cells of the villus, where they are reassembled into fats/lipids and then enter the lacteal, which drains into the lymphatic system.

(b) (i) Diffusion is a passive process down a concentration gradient (high to low) not requiring energy. Active transport is an active process against a concentration gradient (low to high) using energy from respiration and transport/carrier proteins.
(ii) Hepatic portal vein.

(c) Respiration to produce energy, conversion to glycogen for storage (glycogenesis), and conversion to fat/lipids for storage.

评分标准

(a) (i) (max 4 marks from:)
1. Villi/microvilli provide a large surface area;
2. Single-cell-thick epithelium / thin wall ensures a short diffusion distance;
3. Extensive capillary network maintains a steep concentration gradient;
4. Lacteal present in the centre of each villus for lipid transport;
5. Epithelial cells have many mitochondria to provide energy for active transport;
[4]

(a) (ii)
1. Diffuse into epithelial cells (re-forming into lipids);
2. Pass into the lacteal;
[2]

(b) (i) (max 3 marks from:)
1. Diffusion is down a concentration gradient AND active transport is against a concentration gradient;
2. Diffusion does not require energy (passive) AND active transport requires energy (active);
3. Active transport requires carrier/transport proteins (in cell membrane);
[3]

(b) (ii)
1. Hepatic portal vein; [1]

(c) (max 3 marks from:)
1. Respiration / broken down for energy;
2. Converted to glycogen (for storage);
3. Converted to fat/lipids;
4. Released into the general circulation to maintain blood glucose levels;
[3]
题目 4 · structured
14
4 (a) Human activities have led to many species becoming endangered.

(i) State three reasons why a species may become endangered.

(ii) Explain how captive breeding programmes can help conserve endangered animal species.

(b) Plant species are also at risk of extinction and are conserved using seed banks.

(i) Describe the role of seed banks in conserving plant biodiversity.

(ii) Explain why seeds are kept in cold, dry conditions in seed banks.

(c) Sustainable development meets the needs of people without damage to the environment.

Identify two resources that can be managed sustainably.
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解题

(a) (i) Habitat destruction, overhunting/poaching, pollution, introduction of invasive species, or climate change.
(ii) Captive breeding provides a safe environment free from predators/poachers. Organisms are given medical care and optimal nutrition to increase reproductive success. Genetic diversity is managed via studbooks to prevent inbreeding. Individuals can then be reintroduced into their natural habitats.

(b) (i) Seed banks store seeds from a wide variety of plant species to preserve genetic diversity and prevent extinction. If a plant species becomes extinct in the wild, seeds can be germinated to restore the species.
(ii) Cold and dry conditions prevent decay by bacteria/fungi and reduce the metabolic activity of the seeds, keeping them dormant and viable for long periods.

(c) Forests (timber) and fish stocks.

评分标准

(a) (i) (max 3 marks from:)
1. Habitat loss / deforestation / urbanization;
2. Hunting / poaching / overexploitation;
3. Climate change / rising temperatures;
4. Pollution (pesticides, heavy metals, plastic);
5. Invasive/introduced species (competition or predation);
6. Disease;
[3]

(a) (ii) (max 4 marks from:)
1. Provides a safe environment free from predators/poaching;
2. Nutritional support/medical care to increase survival;
3. Use of IVF / artificial insemination;
4. Selection of mates to maintain genetic diversity / reduce inbreeding;
5. Reintroduction into protected wild habitats;
[4]

(b) (i) (max 3 marks from:)
1. Stores a wide variety of seeds/species to preserve biodiversity;
2. Acts as a backup in case of wild extinction;
3. Saves space/is cost-effective compared to conserving whole ecosystems;
4. Maintains genetic variations for future crop breeding/research;
[3]

(b) (ii)
1. Prevents/slows down germination / keeps seeds dormant;
2. Prevents growth of decomposers/bacteria/fungi / prevents decay;
[2]

(c)
1. Forests / timber;
2. Fish stocks / marine resources;
(Accept other valid examples of manageable natural resources) [2]
题目 5 · structured
13
5 (a) Microorganisms are widely used in biotechnology.

(i) State two reasons why bacteria and fungi are suitable for use in biotechnology.

(ii) Write the balanced chemical equation for anaerobic respiration in yeast.

(b) Biofuels can be produced using plant materials and yeast.

(i) Explain how yeast is used to produce ethanol for biofuel.

(ii) Discuss one environmental advantage and one environmental disadvantage of using biofuels instead of fossil fuels.

(c) Pectinase is an enzyme used in the production of fruit juice.

(i) State the substrate and the product of the reaction catalysed by pectinase.

(ii) Explain why heating pectinase to high temperatures before adding it to fruit pulp would prevent it from functioning.
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解题

(a) (i) They reproduce rapidly, can grow on cheap nutrients/substrates, contain plasmids for genetic engineering, and do not present the ethical issues associated with using animals.
(ii) \(C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2\)

(b) (i) Plant material is broken down to release simple sugars. Yeast respires these sugars anaerobically (fermentation) to produce ethanol and carbon dioxide. The ethanol is then separated/purified by distillation.
(ii) Advantage: Biofuels are renewable/carbon-neutral as plants absorb carbon dioxide during photosynthesis. Disadvantage: Large areas of land are needed to grow fuel crops, leading to habitat destruction/deforestation or food shortages.

(c) (i) Substrate: pectin. Product: soluble sugars/galacturonic acid (accept simple sugars).
(ii) High temperatures cause the enzyme to denature. The active site changes shape permanently, meaning the substrate (pectin) can no longer fit/bind to it, and no enzyme-substrate complexes can form.

评分标准

(a) (i) (max 2 marks from:)
1. Rapid rate of growth / reproduction;
2. Simple/cheap nutritional requirements;
3. No ethical concerns (compared to animals);
4. Share the same genetic code / contain plasmids for easy genetic modification;
5. Ability to make complex molecules;
[2]

(a) (ii)
1. Correct formula for glucose on left (\(C_6H_{12}O_6\)) AND correct formula for ethanol (\(C_2H_5OH\)) and carbon dioxide (\(CO_2\)) on right;
2. Correctly balanced: \(C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2\);
[2]

(b) (i) (max 3 marks from:)
1. Plant matter is crushed / broken down to release glucose/sugars;
2. Yeast respires anaerobically / ferments the sugars;
3. In the absence of oxygen;
4. Distillation is used to purify/concentrate the ethanol;
[3]

(b) (ii)
1. Advantage: Carbon-neutral / renewable / reduces greenhouse gas emissions (since plants absorb \(CO_2\) during growth);
2. Disadvantage: Monoculture / habitat destruction to clear land for biofuel crops / uses land that could grow food;
[2]

(c) (i)
1. Substrate: pectin;
2. Product: soluble sugars / galacturonic acid (ignore glucose);
[2]

(c) (ii)
1. Enzyme is denatured / active site changes shape;
2. Substrate no longer fits / cannot bind (no enzyme-substrate complexes form);
[2]
题目 6 · structured
13
6 (a) In humans, the ABO blood group system is determined by three alleles: \(I^A\), \(I^B\), and \(I^O\).

(i) Explain what is meant by the term codominance, using blood groups as an example.

(ii) A man with blood group A (heterozygous) and a woman with blood group B (heterozygous) have a child.

Construct a genetic diagram to show the possible genotypes and phenotypes of their offspring.

(b) Sickle-cell anaemia is another genetic condition.

(i) Describe the symptoms of sickle-cell anaemia.

(ii) Explain why the allele for sickle-cell anaemia is maintained in human populations in areas where malaria is common.
查看答案详解

解题

(a) (i) Codominance is when both alleles are expressed in the phenotype of a heterozygote. For example, individuals with the genotype \(I^AI^B\) express both A and B antigens on their red blood cells, resulting in blood group AB.
(ii) Parental genotypes: Man is \(I^AI^O\), Woman is \(I^BI^O\).
Gametes: Man: \(I^A\) and \(I^O\); Woman: \(I^B\) and \(I^O\).
Offspring Genotypes: \(I^AI^B\), \(I^BI^O\), \(I^AI^O\), \(I^OI^O\).
Offspring Phenotypes: Blood group AB, Blood group B, Blood group A, Blood group O (ratio 1:1:1:1).

(b) (i) Abnormal crescent-shaped red blood cells, reduced oxygen transport causing fatigue/weakness, cells block small blood capillaries causing joint pain/organ damage, and anaemia due to rapid destruction of red blood cells.
(ii) Heterozygous individuals (\(Hb^AHb^S\)) have the sickle-cell trait and are highly resistant to malaria. In areas where malaria is common, these heterozygotes have a survival advantage over homozygotes for normal haemoglobin (who may die of malaria) and homozygotes for sickle-cell anaemia (who suffer from the severe disease). This selective advantage keeps both alleles in the gene pool.

评分标准

(a) (i)
1. Both alleles are expressed in the phenotype (of a heterozygote);
2. Example: \(I^A\) and \(I^B\) alleles produce blood group AB;
[2]

(a) (ii)
1. Correct parental genotypes: \(I^AI^O\) and \(I^BI^O\);
2. Correct gametes: \(I^A\), \(I^O\) and \(I^B\), \(I^O\);
3. Correct offspring genotypes: \(I^AI^B\), \(I^BI^O\), \(I^AI^O\), \(I^OI^O\);
4. Correct phenotypes linked to genotypes: blood group AB, B, A, O;
5. Correct phenotypic ratio: 1:1:1:1 (or 25% each);
[5]

(b) (i) (max 3 marks from:)
1. Red blood cells become sickle/crescent-shaped;
2. Reduced oxygen-carrying capacity / fatigue / shortness of breath / anaemia;
3. Blockage of capillaries/vessels;
4. Pain / swelling in joints / damage to organs;
[3]

(b) (ii) (max 3 marks from:)
1. Heterozygotes (\(Hb^AHb^S\)) have a selective advantage;
2. They are resistant to malaria / malaria parasite cannot easily reproduce in their red blood cells;
3. Homozygous dominant (\(Hb^AHb^A\)) may die from malaria;
4. Homozygous recessive (\(Hb^SHb^S\)) suffer from severe sickle-cell anaemia;
5. Thus, the allele is passed on / maintained in the population by surviving heterozygotes;
[3]

Paper 61 (Alternative to Practical)

Answer all questions. Use a black or dark blue pen. Show your working where required.
2 题目 · 40
题目 1 · Practical
20
**Answer all questions. Use a black or dark blue pen. Show your working where required.**

**1** A student investigated the effect of temperature on the rate of anaerobic respiration in yeast.

They used the following method:
- Set up a water bath at a constant temperature.
- Mix $10\text{ cm}^3$ of a $5\%$ active yeast suspension with $10\text{ cm}^3$ of a $10\%$ glucose solution in a reaction tube.
- Add a $2\text{ mm}$ layer of liquid paraffin (oil) on top of the mixture.
- Connect the reaction tube via a delivery tube to a gas syringe.
- Leave the tube in the water bath for 5 minutes to equilibrate, then record the volume of gas collected after exactly 10 minutes.
- Repeat the experiment at different temperatures: $20^\circ\text{C}$, $30^\circ\text{C}$, $40^\circ\text{C}$, $50^\circ\text{C}$, and $60^\circ\text{C}$.

The student recorded the following volume of gas collected after 10 minutes:
- At $20^\circ\text{C}$: $3.4\text{ cm}^3$
- At $30^\circ\text{C}$: $7.2\text{ cm}^3$
- At $40^\circ\text{C}$: $11.6\text{ cm}^3$
- At $50^\circ\text{C}$: $4.8\text{ cm}^3$
- At $60^\circ\text{C}$: $0.8\text{ cm}^3$

**(a) (i)** Prepare a table to record these results. [4]

**(a) (ii)** Plot a line graph on a grid to show the effect of temperature on the volume of gas collected. [4]

**(a) (iii)** State the temperature at which the rate of anaerobic respiration was the highest. With reference to your graph and enzyme theory, explain the difference in gas production between $40^\circ\text{C}$ and $60^\circ\text{C}$. [3]

**(b) (i)** State the purpose of adding the liquid paraffin layer on top of the yeast-glucose mixture. [1]

**(b) (ii)** Identify the independent variable and the dependent variable in this investigation. [2]

**(c)** Plan an investigation to determine the effect of glucose concentration on the rate of anaerobic respiration in yeast. [6]
查看答案详解

解题

**(a) (i)**
| Temperature / $^\circ\text{C}$ | Volume of gas collected after 10 minutes / $\text{cm}^3$ |
| :---: | :---: |
| 20 | 3.4 |
| 30 | 7.2 |
| 40 | 11.6 |
| 50 | 4.8 |
| 60 | 0.8 |

**(a) (ii)**
- **Axes:** x-axis labeled 'Temperature / $^\circ\text{C}$', y-axis labeled 'Volume of gas collected / $\text{cm}^3$'.
- **Scale:** Linear, appropriate scale occupying at least half of the grid in both directions.
- **Plotting:** All 5 points plotted accurately (within half a small square).
- **Line:** Points joined with a clean line (either ruled straight lines point-to-point or a smooth curve).

**(a) (iii)**
- Highest rate of respiration: $40^\circ\text{C}$.
- At $40^\circ\text{C}$, the temperature is near the optimum for yeast respiratory enzymes, resulting in high kinetic energy, rapid collisions, and a high rate of reaction.
- At $60^\circ\text{C}$, the high temperature has denatured the enzymes; the active site has changed shape, preventing glucose substrate binding, which dramatically decreases gas production.

**(b) (i)** To prevent oxygen from entering the mixture / to maintain anaerobic conditions.

**(b) (ii)**
- Independent variable: Temperature.
- Dependent variable: Volume of gas collected (after 10 minutes).

**(c)**
- **Independent variable:** Prepare at least 5 different concentrations of glucose solution (e.g., $1\%$, $2\%$, $4\%$, $8\%$, $10\%$).
- **Dependent variable:** Measure the volume of gas produced in a set time (e.g., 10 minutes) using a gas syringe.
- **Control variables:** Keep temperature constant using a thermostatically controlled water bath; keep the volume and concentration of yeast suspension constant; keep the total volume of reaction mixture constant.
- **Replication:** Repeat each glucose concentration at least 3 times to calculate a mean and identify anomalies.
- **Safety:** Wear safety goggles and lab coat; handle glassware with care.

评分标准

**(a)(i)**
1. Table drawn with clear boundary lines and separate columns [1]
2. Headings with appropriate units: Temperature / $^\circ\text{C}$ AND Volume / $\text{cm}^3$ [1]
3. All 5 temperatures recorded in order [1]
4. Correct values entered: 3.4, 7.2, 11.6, 4.8, 0.8 [1]

**(a)(ii)**
1. Both axes correctly labeled with units [1]
2. Suitable linear scales using at least half of the grid [1]
3. All 5 points plotted correctly to within half a small square [1]
4. Clean line connecting all plotted points [1]

**(a)(iii)**
1. Correct temperature identified: $40^\circ\text{C}$ [1]
2. Explanation for $40^\circ\text{C}$ (enzymes active / high kinetic energy / optimum temperature) [1]
3. Explanation for $60^\circ\text{C}$ (enzymes denatured / shape of active site changed / no reaction) [1]

**(b)(i)**
1. Exclude oxygen / ensure anaerobic conditions [1]

**(b)(ii)**
1. Independent variable: temperature [1]
2. Dependent variable: volume of gas collected / rate of respiration [1]

**(c)**
1. Prepare at least 5 different concentrations of glucose [1]
2. Method of measuring gas volume in a set time period [1]
3. Key control variable 1: keep temperature constant [1]
4. Key control variable 2: keep yeast suspension volume/concentration constant [1]
5. Replications: repeat the procedure at least three times per concentration to calculate a mean [1]
6. Safety precaution: use of goggles / safe heating method described [1]
题目 2 · Practical
20
**2** A student investigated the density of stomata on the leaves of a privet plant (*Ligustrum vulgare*).

They applied clear nail varnish to a small area of both the upper and lower surfaces of a leaf. Once dry, they peeled off the varnish using clear tape and placed it on a microscope slide to create a replica of the epidermis. The replica was viewed under a microscope at $\times 400$ magnification, and the number of stomata in three different fields of view was counted for each surface.

Their counts were as follows:
- **Upper surface epidermis replica:**
- Field 1: 1 stoma
- Field 2: 2 stomata
- Field 3: 0 stomata

- **Lower surface epidermis replica:**
- Field 1: 13 stomata
- Field 2: 15 stomata
- Field 3: 14 stomata

**(a) (i)** Calculate the mean number of stomata for both the upper and lower surfaces of the leaf. Show your working. [2]

**(a) (ii)** State a conclusion based on these results. [1]

**(a) (iii)** Explain how the difference in stomatal distribution between the upper and lower surfaces is an adaptation that helps land plants survive. [2]

**(b) (i)** At $\times 400$ magnification, the field of view of the microscope is a circle with a diameter of $0.46\text{ mm}$.
Calculate the area of this field of view in $\text{mm}^2$.
Use the formula:
$$\text{Area} = \pi r^2$$
where $\pi = 3.14$ and $r$ is the radius of the field of view.
Show your working. Give your answer to two significant figures. [3]

**(b) (ii)** Using your calculated area from **(b)(i)** and the mean number of stomata on the lower surface from **(a)(i)**, estimate the stomatal density (number of stomata per $\text{mm}^2$) on the lower surface of the leaf. Show your working. [2]

**(c) (i)** Draw a large, clear diagram of a single stoma as seen under a microscope: it consists of a central pore surrounded by two bean-shaped guard cells, each containing several small circular chloroplasts and a thicker cell wall on the side facing the pore. [4]

**(c) (ii)** Label one guard cell and the stomatal pore on your drawing. [2]

**(d)** Describe how you would test a leaf to show that it has been photosynthesising by testing for the presence of starch. [4]
查看答案详解

解题

**(a) (i)**
- **Upper surface mean:** $\frac{1 + 2 + 0}{3} = 1.0$
- **Lower surface mean:** $\frac{13 + 15 + 14}{3} = 14.0$

**(a) (ii)** Stomata are significantly more numerous/denser on the lower surface of the leaf than on the upper surface.

**(a) (iii)** The upper surface is exposed to direct sunlight and wind, which increases the rate of evaporation. Having fewer stomata on the upper surface reduces water loss by transpiration, helping the plant conserve water.

**(b) (i)**
- Diameter = $0.46\text{ mm}$, so radius $r = 0.23\text{ mm}$.
- $\text{Area} = 3.14 \times (0.23)^2 = 3.14 \times 0.0529 = 0.166106\text{ mm}^2$.
- Rounding to two significant figures gives: $0.17\text{ mm}^2$.

**(b) (ii)**
- Mean count on lower surface = $14.0$
- $\text{Stomatal Density} = \frac{\text{Mean count}}{\text{Area}} = \frac{14.0}{0.166106} = 84.28\text{ stomata per mm}^2$ (or $\frac{14.0}{0.17} = 82.35\text{ stomata per mm}^2$). Accept values in the range of $82\text{ to }85$.

**(c) (i)**
- **Drawing criteria:** Clean, single continuous lines with no sketchy/feathered outlines or shading. The drawing should occupy at least half the available space. Two bean-shaped guard cells should be shown flanking a central open pore. The cell walls on the side facing the pore must be drawn visibly thicker than the outer walls. Inside each guard cell, multiple small circles should represent chloroplasts.

**(c) (ii)** Label lines must point directly and unambiguously to:
- **Guard cell:** One of the bean-shaped cells.
- **Stomatal pore:** The central opening/gap.

**(d)**
1. Place the leaf in boiling water for about 1 minute to break the cell walls and kill the cells.
2. Place the boiled leaf in hot ethanol using an electric water bath (no naked flames, as ethanol is highly flammable) to extract the chlorophyll and decolourise the leaf.
3. Dip the leaf briefly in warm water to soften it.
4. Spread the leaf flat on a white tile and add a few drops of iodine solution. A color change from brown to blue-black indicates starch is present.

评分标准

**(a)(i)**
1. Correct calculation of upper mean (1.0) and lower mean (14.0) [1]
2. Working shown [1]

**(a)(ii)**
1. Correct conclusion stating there are more stomata on the lower surface than the upper surface [1]

**(a)(iii)**
1. Upper surface is hotter / more exposed to sunlight [1]
2. Less transpiration / reduces water loss [1]

**(b)(i)**
1. Correct identification of radius ($0.23\text{ mm}$) [1]
2. Correct substitution into formula: $3.14 \times (0.23)^2$ [1]
3. Final answer of $0.17\text{ mm}^2$ (must be 2 s.f. and have units) [1]

**(b)(ii)**
1. Division of mean count (14.0) by area from (b)(i) [1]
2. Correct calculation within range 82 to 85 [1]

**(c)(i)**
1. Large diagram drawn with single, clean continuous lines and no shading [1]
2. Bean-shaped guard cells surrounding an open pore [1]
3. Inner walls of guard cells drawn thicker than outer walls [1]
4. Chloroplasts shown as small circular organelles inside guard cells [1]

**(c)(ii)**
1. Accurate label line to guard cell [1]
2. Accurate label line to stomatal pore [1]

**(d)**
1. Boil leaf in water (kills leaf / breaks cell walls) [1]
2. Heat leaf in ethanol using a water bath (removes chlorophyll) [1]
3. Safety: mention electric water bath / no naked flames [1]
4. Spread on tile, add iodine solution, blue-black color indicates starch [1]

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