- A.the number of protein molecules in their bodies
- B.the sequence of bases in their DNA
- C.the exact length of their wings
- D.the types of habitats they occupy
Cambridge IGCSE · thinka 原创模拟试题
2025 Cambridge IGCSE Biology (0610) 模拟试题及答案详解
Paper 21 (Extended 選擇題)
- A.\(\times\) 250
- B.\(\times\) 400
- C.\(\times\) 2500
- D.\(\times\) 90000
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- A.low humidity, high temperature, high wind speed
- B.low humidity, low temperature, low wind speed
- C.high humidity, high temperature, high wind speed
- D.high humidity, low temperature, low wind speed
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- A.Oxygenated and deoxygenated blood mix in the heart to lower pressure.
- B.Blood is pumped at a higher pressure to the body tissues.
- C.Red blood cells can transport larger amounts of carbon dioxide.
- D.Less energy from respiration is required to pump blood through the lungs.
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- A.1 and 2
- B.1 and 3
- C.2 and 4
- D.3 and 4
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- A.active transport
- B.deamination
- C.denitrification
- D.nitrification
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- A.fertilizer runoff -> algal bloom -> death of plants -> decomposers respire -> oxygen depletion
- B.fertilizer runoff -> decomposers respire -> death of plants -> algal bloom -> oxygen depletion
- C.oxygen depletion -> algal bloom -> decomposers respire -> fertilizer runoff -> death of plants
- D.algal bloom -> fertilizer runoff -> death of plants -> oxygen depletion -> decomposers respire
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- A.the original DNA strand in the nucleus
- B.a molecule of messenger RNA (mRNA)
- C.a strand of double-stranded DNA in the cytoplasm
- D.an enzyme in the cytoplasm
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- A.high temperature, high wind speed, high light intensity, low humidity
- B.high temperature, low wind speed, low light intensity, high humidity
- C.low temperature, high wind speed, high light intensity, low humidity
- D.low temperature, low wind speed, low light intensity, high humidity
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- A.Petals are large and brightly coloured to protect the reproductive organs.
- B.Stigmas are long and feathery to catch wind-borne pollen.
- C.Nectaries are located at the base of the flower to reward visiting insects.
- D.Anthers are dangling outside the flower to release pollen into the wind.
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- A.Blood can be pumped to the lungs at a very high pressure to speed up gas exchange.
- B.Deoxygenated blood and oxygenated blood mix in the heart, increasing metabolic efficiency.
- C.Blood pressure can be kept high in the systemic circulation to deliver oxygen rapidly to tissues, while remaining lower in the pulmonary circulation to protect lung capillaries.
- D.Blood flows through the heart only once during each complete circuit of the body, reducing energy expenditure.
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- A.Toxic chemicals released directly by agricultural fertilisers.
- B.Increased concentration of dissolved carbon dioxide produced by rapidly growing algae.
- C.A lack of dissolved oxygen in the water due to aerobic respiration by decomposers.
- D.Blockage of the fish gills by the dense layer of floating algae.
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1. *Canis lupus*
2. *Canis latrans*
3. *Felis catus*
Which statement about these organisms is correct?
- A.Organisms 1 and 2 belong to different genera.
- B.Organisms 1 and 2 share more recent common ancestry with each other than with organism 3.
- C.Organisms 2 and 3 belong to the same species.
- D.Organisms 1, 2 and 3 all belong to the same genus.
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What is the magnification of the photomicrograph?
- A.\(\times 30\)
- B.\(\times 300\)
- C.\(\times 3\text{ }000\)
- D.\(\times 30\text{ }000\)
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First, convert the measurements to the same units.
Image size = \( 45\text{ mm} = 45\text{ }000\text{ }\mu\text{m} \).
Actual size = \( 1.5\text{ }\mu\text{m} \).
\(\text{Magnification} = \frac{45\text{ }000\text{ }\mu\text{m}}{1.5\text{ }\mu\text{m}} = \times 30\text{ }000\).
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- A.volume of thorax decreases, pressure in lungs decreases
- B.volume of thorax decreases, pressure in lungs increases
- C.volume of thorax increases, pressure in lungs decreases
- D.volume of thorax increases, pressure in lungs increases
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- A.active transport by root hair cells
- B.denitrification by anaerobic bacteria
- C.nitrification by nitrifying bacteria
- D.nitrogen fixation by bacteria in root nodules
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- A.0.7 \(\mu\text{m}\)
- B.7.0 \(\mu\text{m}\)
- C.70.0 \(\mu\text{m}\)
- D.112.0 \(\mu\text{m}\)
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- A.warm, humid, still air
- B.cool, dry, moving air
- C.warm, dry, moving air
- D.warm, dry, still air
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- A.Fish have a double circulatory system, whereas mammals have a single circulatory system.
- B.Fish hearts have a septum to completely separate oxygenated and deoxygenated blood, whereas mammalian hearts do not.
- C.In fish, blood passes through the heart once for each complete circuit of the body, whereas in mammals it passes through twice.
- D.Mammals pump blood under lower pressure to the body tissues than fish to prevent capillary damage.
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- A.insect-pollinated; heavy, spiky pollen grains
- B.wind-pollinated; light, smooth pollen grains
- C.wind-pollinated; large, sticky pollen grains
- D.insect-pollinated; small, smooth pollen grains
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- A.X is nitrification; Y is denitrification
- B.X is nitrogen fixation; Y is nitrification
- C.X is deamination; Y is denitrification
- D.X is nitrification; Y is assimilation
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- A.to locate and isolate the human insulin gene sequence from human chromosomes
- B.to cut open the circular bacterial plasmid at a specific restriction site
- C.to join the isolated human insulin gene to the cut plasmid DNA
- D.to stimulate the rapid binary fission of the transformed bacteria
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- A.It delivers specific amino acids to the nucleus to be joined together.
- B.It carries a copy of the gene sequence from the nucleus to the ribosome.
- C.It stays in the nucleus to act as a permanent master template for replication.
- D.It acts as a structural component of the nuclear membrane.
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- A.sweat glands increase secretion; arterioles dilate
- B.sweat glands increase secretion; arterioles constrict
- C.sweat glands decrease secretion; arterioles dilate
- D.sweat glands decrease secretion; arterioles constrict
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- A.In a single circulatory system, blood passes through the heart twice for each complete circuit.
- B.In a double circulatory system, the pressure of blood going to the lungs is higher than the pressure of blood going to the body.
- C.A double circulatory system maintains a higher blood pressure to the body tissues, leading to more efficient oxygen delivery.
- D.A single circulatory system contains a four-chambered heart, which prevents any mixing of oxygenated and deoxygenated blood.
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The distance moved by the air bubble in 5 minutes was measured under two different conditions:
- Condition 1: \(15^\circ\text{C}\), high humidity, still air
- Condition 2: \(30^\circ\text{C}\), low humidity, windy
Which distances moved by the bubble would be expected for each condition?
- A.Condition 1: \(5\text{ mm}\); Condition 2: \(5\text{ mm}\)
- B.Condition 1: \(5\text{ mm}\); Condition 2: \(62\text{ mm}\)
- C.Condition 1: \(62\text{ mm}\); Condition 2: \(5\text{ mm}\)
- D.Condition 1: \(62\text{ mm}\); Condition 2: \(62\text{ mm}\)
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- A.large and brightly coloured petals; feathery stigmas hanging outside; small, light and smooth pollen
- B.small and green petals; feathery stigmas hanging outside; small, light and smooth pollen
- C.small and green petals; sticky stigmas inside the flower; large, heavy and sticky pollen
- D.large and brightly coloured petals; sticky stigmas inside the flower; large, heavy and sticky pollen
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1. Algae grow rapidly on the surface of the water.
2. Decomposers multiply and respire aerobically.
3. Light is blocked from reaching plants on the river bed.
4. Run-off of fertiliser into the river occurs.
5. Fish die due to a lack of dissolved oxygen.
What is the correct sequence of these events?
- A.4 \(\rightarrow\) 1 \(\rightarrow\) 3 \(\rightarrow\) 2 \(\rightarrow\) 5
- B.4 \(\rightarrow\) 3 \(\rightarrow\) 1 \(\rightarrow\) 2 \(\rightarrow\) 5
- C.1 \(\rightarrow\) 4 \(\rightarrow\) 3 \(\rightarrow\) 5 \(\rightarrow\) 2
- D.1 \(\rightarrow\) 3 \(\rightarrow\) 4 \(\rightarrow\) 2 \(\rightarrow\) 5
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What is the magnification of the micrograph?
- A.\(\times 20\)
- B.\(\times 200\)
- C.\(\times 2000\)
- D.\(\times 20\,000\)
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解题
\(30\text{ mm} = 30 \times 1000\ \mu\text{m} = 30\,000\ \mu\text{m}\).
Using the formula:
\(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{30\,000\ \mu\text{m}}{1.5\ \mu\text{m}} = 20\,000\).
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- A.external intercostal muscles contract; internal intercostal muscles relax; diaphragm contracts; volume of thorax increases
- B.external intercostal muscles relax; internal intercostal muscles contract; diaphragm relaxes; volume of thorax decreases
- C.external intercostal muscles contract; internal intercostal muscles relax; diaphragm relaxes; volume of thorax decreases
- D.external intercostal muscles relax; internal intercostal muscles contract; diaphragm contracts; volume of thorax increases
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- A.denitrification, which increases the concentration of nitrate ions in the soil
- B.nitrification, which converts nitrogen gas in the air directly into ammonia
- C.nitrogen fixation, which converts inert nitrogen gas into nitrogen-containing compounds
- D.decomposition, which converts nitrate ions in the soil into proteins in root nodules
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- A.DNA molecules are proteins that contain genes, which are made of different amino acids.
- B.A gene is a specific sequence of bases in a DNA molecule that codes for the structure of a protein.
- C.Proteins are translated inside the nucleus to form mRNA, which then codes for DNA replication.
- D.Each chromosome contains a single gene that codes for all the different proteins in a cell.
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- A.× 24
- B.× 240
- C.× 2 400
- D.× 24 000
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\[ \text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} \]
First, convert the image length into the same units as the actual size (micrometres, \( \mu\text{m} \)):
\[ 3.6\text{ cm} = 36\text{ mm} = 36\,000\ \mu\text{m} \]
Next, calculate the magnification:
\[ \text{Magnification} = \frac{36\,000\ \mu\text{m}}{15\ \mu\text{m}} = 2\,400 \]
Therefore, the magnification is \( \times 2\,400 \).
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- A.The stomata close immediately to prevent any water loss.
- B.The concentration gradient of water vapour between the inside of the leaf and the external air is reduced.
- C.The rate of evaporation from the surfaces of the mesophyll cells is increased.
- D.The cohesion between water molecules in the xylem vessels is disrupted.
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- A.Blood can be pumped to the respiring tissues at a higher pressure.
- B.Oxygenated and deoxygenated blood are mixed to maintain a constant body temperature.
- C.The blood flows more slowly through the systemic circulation than the pulmonary circulation.
- D.Fewer heart valves are required to control the direction of blood flow.
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- A.deamination
- B.denitrification
- C.nitrification
- D.nitrogen fixation
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1 They have a rapid rate of reproduction.
2 They contain plasmids that can be easily manipulated.
3 They share the same genetic code as other organisms.
4 Their cells contain membrane-bound nuclei.
- A.1, 2 and 3
- B.1, 2 and 4
- C.1, 3 and 4
- D.2, 3 and 4
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- A.mRNA is a double-stranded molecule that remains inside the nucleus.
- B.mRNA carries a copy of the gene's base sequence from the nucleus to a ribosome.
- C.mRNA directly codes for the production of starch molecules.
- D.mRNA acts as an enzyme to catalyse the bonding of amino acids.
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- A.arterioles constrict, sweat production decreases
- B.arterioles constrict, sweat production increases
- C.arterioles dilate, sweat production decreases
- D.arterioles dilate, sweat production increases
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1. small, green petals
2. long filaments that hang outside the flower
3. large, feathery stigmas
Which method of pollination is this flower adapted for, and what is the reason?
- A.insect pollination, because the feathery stigmas provide a landing platform for bees
- B.insect pollination, because the green petals camouflage the flower from herbivores
- C.wind pollination, because the feathery stigmas can easily trap drifting pollen grains
- D.wind pollination, because the long filaments protect the ovules from wind damage
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Paper 41 (Extended Theory)
1 A student investigated the rate of transpiration of a leafy shoot under different wind conditions using a potometer.
(a) Define the term transpiration. [2]
(b) (i) Explain the function of the reservoir in a potometer. [2]
(ii) State two precautions that must be taken when setting up a potometer to ensure accurate results. [2]
(c) Table 1.1 shows the results of the student's investigation under three different wind speeds.
Table 1.1
• Calm air: 1.5 mm/min
• Gentle breeze: 3.8 mm/min
• Strong wind: 6.4 mm/min
Describe and explain the effect of wind speed on the rate of transpiration shown in Table 1.1. [5]
(d) State two environmental factors, other than wind speed, that affect the rate of transpiration. [2]
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(a) Transpiration is the loss of water vapour from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by the diffusion of water vapour through the stomata.
(b) (i) The reservoir allows the air bubble to be pushed back to the start of the capillary tube (scale) so that the measurement can be repeated.
(ii) 1. Cut the stem of the shoot under water to prevent air from entering the xylem.
2. Seal all joints of the potometer apparatus with petroleum jelly to ensure it is completely airtight.
(c) Description: As the wind speed increases, the rate of transpiration increases (from 1.5 mm/min in calm air to 6.4 mm/min in strong wind).
Explanation: Moving air blows away the water vapour that accumulates on the surface of the leaf (boundary layer). This maintains a steep water vapour concentration gradient between the inside of the leaf air spaces and the outside atmosphere, leading to faster diffusion of water vapour out through the stomata.
(d) Any two from: temperature, humidity, light intensity.
评分标准
(a) max [2]:
- loss of water vapour from leaves / aerial parts of plant [1]
- by evaporation from surfaces of mesophyll cells [1]
- followed by diffusion of water vapour through stomata [1]
(b) (i) max [2]:
- to return the air bubble to the start / zero mark [1]
- allows repeats / multiple measurements to be taken [1]
(b) (ii) max [2]:
- cut shoot under water [1]
- seal joints with Vaseline / grease / petroleum jelly [1]
- ensure leaves are dry before starting [1]
(c) max [5]:
- rate increases as wind speed increases / describe trend with data quotes [1]
- wind removes the saturated layer of air / boundary layer / water vapour outside the leaf [1]
- this maintains / increases the water vapour concentration gradient [1]
- between the air spaces inside the leaf and the external air [1]
- leading to faster diffusion of water vapour through stomata [1]
(d) max [2]:
- temperature [1]
- humidity [1]
- light intensity [1]
2 (a) Contrast the structural features of insect-pollinated and wind-pollinated flowers. Complete the table by describing the petals, stigmas, and pollen grains for each type of flower. [3]
(b) A pollen grain from an insect-pollinated flower has an actual diameter of 80 micrometres. In a diagram, the drawn diameter of this pollen grain is 40 millimetres (mm).
(i) State the formula used to calculate magnification. [1]
(ii) Calculate the magnification of the pollen grain in the diagram. Show your working. [3]
(c) Describe the sequence of events leading to fertilisation in a flowering plant, starting from when a compatible pollen grain lands on the stigma. [6]
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(a)
- Petals: Insect-pollinated are large and brightly coloured; wind-pollinated are small, green or dull-coloured.
- Stigmas: Insect-pollinated are sticky and enclosed within the flower; wind-pollinated are long, feathery and hang outside the flower.
- Pollen: Insect-pollinated are larger, sticky or spiky, produced in smaller quantities; wind-pollinated are small, light, smooth, produced in massive quantities.
(b) (i) Magnification = Image size (size of drawing) / Actual size
(b) (ii) Convert units: 40 mm = 40 * 1000 = 40000 micrometres.
Magnification = 40000 / 80 = x500.
(c) 1. The pollen grain germinates on the sticky stigma.
2. A pollen tube grows down through the style towards the ovary.
3. The growth of the pollen tube is controlled by the pollen tube nucleus.
4. The pollen tube enters the ovary and reaches an ovule, entering through the micropyle.
5. The male gamete nucleus travels down the pollen tube.
6. The male nucleus fuses with the female gamete nucleus (egg cell nucleus) inside the ovule to form a diploid zygote.
评分标准
(a) max [3]:
- Petals: large/brightly coloured vs small/dull/green [1]
- Stigmas: sticky/inside flower vs feathery/hanging outside flower [1]
- Pollen: sticky/spiky/large/few vs smooth/light/small/many [1]
(b) (i) [1]:
- Magnification = Image size / Actual size [1]
(b) (ii) max [3]:
- conversion of 40 mm to 40000 micrometres [1]
- correct substitution: 40000 / 80 [1]
- correct calculation: x500 or 500 [1]
(c) max [6]:
- pollen grain germinates / produces pollen tube [1]
- pollen tube grows down the style [1]
- enters the ovary / enters the ovule [1]
- through the micropyle [1]
- male nucleus / gamete travels down the tube [1]
- male and female nuclei fuse / join [1]
- to form a zygote / diploid cell [1]
3 Mammals possess a closed, double circulatory system, whereas fish have a single circulatory system.
(a) Explain what is meant by a double circulatory system. [2]
(b) Explain the physiological advantages of a double circulatory system in mammals compared to the single circulatory system of fish. [4]
(c) (i) Describe two structural differences between a typical human artery and a human vein. [2]
(ii) Explain how the structure of an artery is adapted to its function of carrying blood away from the heart. [4]
(d) State the name of the blood vessel that: [2]
(i) carries deoxygenated blood to the lungs.
(ii) carries oxygenated blood to the kidneys.
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(a) A double circulatory system means that for every complete circuit of the body, blood passes through the heart twice (pulmonary circuit and systemic circuit).
(b) 1. Keeps oxygenated and deoxygenated blood separate, preventing mixing and maximizing oxygen delivery to tissues.
2. Allows blood to be pumped at a higher pressure to body tissues, ensuring faster delivery of oxygen and glucose for aerobic respiration.
3. Blood to the lungs can be pumped at a lower pressure, preventing damage to the delicate lung capillaries while still allowing efficient gas exchange.
(c) (i) 1. Arteries have thick muscular and elastic walls, whereas veins have thin walls with less muscle and elastic tissue.
2. Arteries have a narrow lumen, whereas veins have a wide lumen (and contain valves to prevent backflow).
(c) (ii) 1. The thick layer of muscle and elastic fibers allows the wall to stretch and recoil to withstand and maintain high blood pressure from the heart's pumping action.
2. The elastic recoil helps to smooth out the pulse of blood.
3. The narrow lumen maintains high blood pressure to transport blood quickly to body tissues.
(d) (i) Pulmonary artery.
(ii) Renal artery.
评分标准
(a) max [2]:
- blood passes through the heart twice [1]
- for one complete circuit / round of the body [1]
(b) max [4]:
- oxygenated and deoxygenated blood are kept separate / do not mix [1]
- blood can be pumped at high pressure to the body [1]
- blood is pumped at a lower pressure to the lungs [1]
- faster / more efficient delivery of oxygen / glucose to respiring tissues [1]
- supports higher metabolic rate / active lifestyle / warm-bloodedness [1]
(c) (i) max [2]:
- Arteries: thicker wall / more muscle / more elastic tissue vs Veins: thinner wall [1]
- Arteries: narrow lumen vs Veins: wide lumen [1]
- Arteries: no valves vs Veins: have valves [1]
(c) (ii) max [4]:
- thick muscular wall withstands high blood pressure [1]
- elastic tissue stretches and recoils [1]
- recoil maintains / regulates blood pressure / smooths blood flow [1]
- narrow lumen maintains high pressure [1]
(d) (i) [1]:
- pulmonary artery [1]
(d) (ii) [1]:
- renal artery [1]
4 Agricultural runoff containing chemical fertilisers is a major cause of pollution in aquatic ecosystems.
(a) Plants require nitrate ions absorbed from fertilisers.
State the name of one major group of biological molecules that plants synthesis using nitrate ions. [1]
(b) Describe and explain the sequence of events that occurs when excess nitrate fertiliser runs off into a lake, leading to eutrophication. [6]
(c) Untreated sewage can also enter waterways.
(i) State the general name given to the group of organisms that decompose organic waste in sewage. [1]
(ii) Explain why untreated sewage discharged into a river causes a rapid decrease in the concentration of dissolved oxygen. [2]
(d) Nitrogen gas in the atmosphere is converted into forms plants can use.
(i) Define the term nitrogen fixation. [2]
(ii) State the name of the process that converts ammonium ions into nitrate ions in the soil. [2]
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解题
(a) Proteins / amino acids.
(b) 1. High nitrate levels cause rapid growth of algae on the water surface (algal bloom).
2. This layer of algae blocks sunlight from reaching submerged aquatic plants.
3. Submerged plants cannot photosynthesise and therefore die.
4. Aerobic bacteria (decomposers) multiply rapidly as they feed on the dead plant matter.
5. These bacteria respire aerobically, using up the dissolved oxygen in the water.
6. The lack of dissolved oxygen causes fish and other aquatic animals to suffocate and die.
(c) (i) Decomposers (or bacteria/fungi).
(c) (ii) Untreated sewage contains high levels of organic matter. Decomposing bacteria feed on this organic matter, multiply rapidly, and consume large amounts of dissolved oxygen through aerobic respiration.
(d) (i) Nitrogen fixation is the process of converting inert atmospheric nitrogen gas into nitrogen-containing compounds, such as ammonia or nitrates, that can be absorbed and used by plants.
(d) (ii) Nitrification.
评分标准
(a) [1]:
- proteins / amino acids [1]
(b) max [6]:
- algal bloom / rapid growth of algae on surface [1]
- blocks sunlight from entering water [1]
- submerged plants cannot photosynthesise [1]
- submerged plants die [1]
- bacteria / decomposers feed on dead plants and multiply [1]
- bacteria respire aerobically [1]
- dissolved oxygen concentration decreases [1]
- fish / aquatic animals suffocate / die [1]
(c) (i) [1]:
- decomposers / bacteria / fungi [1]
(c) (ii) max [2]:
- sewage provides organic matter / food for bacteria [1]
- bacteria undergo aerobic respiration [1]
- using up oxygen / oxygen depletion [1]
(d) (i) max [2]:
- conversion of atmospheric nitrogen gas [1]
- into ammonia / ammonium / nitrates / compounds plants can use [1]
(d) (ii) [2]:
- nitrification [1]
5 Classification systems help scientists identify and organize living organisms.
(a) The binomial system is used to name organisms worldwide.
(i) Define the term binomial system. [2]
(ii) State what the first word and the second word of a binomial name represent. [2]
(b) Vertebrates are divided into five classes.
State two structural features that distinguish amphibians from reptiles. [2]
(c) Modern classification relies heavily on molecular evidence.
Explain why comparing DNA base sequences is a more reliable method of classification than comparing morphological features. [3]
(d) Eukaryotic cells have distinct structures.
(i) State two structures present in a palisade mesophyll cell that are absent in a human cheek cell. [2]
(ii) Outline the function of the cellulose cell wall in plant cells. [2]
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解题
(a) (i) The binomial system is an internationally agreed system in which the scientific name of an organism is made up of two parts: the genus name and the species name.
(a) (ii) First word: Genus.
Second word: Species.
(b) 1. Amphibians have moist, scale-less skin, whereas reptiles have dry skin with scales.
2. Amphibians lay soft, jelly-like eggs in water, whereas reptiles lay eggs with leathery/rubbery shells on land.
(c) 1. Morphological features can be misleading due to convergent evolution, where unrelated organisms evolve similar structures (analogous structures) to adapt to similar environments.
2. DNA sequences provide a direct record of genetic inheritance and evolutionary history.
3. The closer the DNA base sequences are between two species, the more recently they shared a common ancestor.
(d) (i) Cell wall, chloroplasts (or large permanent vacuole).
(d) (ii) The cellulose cell wall is fully permeable and rigid. It resists turgor pressure inside the cell, preventing the cell from bursting when it takes up water by osmosis, and provides structural support to the plant.
评分标准
(a) (i) max [2]:
- internationally agreed system of naming [1]
- using two names / genus and species [1]
(a) (ii) [2]:
- First word: genus [1]
- Second word: species [1]
(b) max [2]:
- skin: moist/no scales vs dry/scales [1]
- reproduction: jelly-like eggs in water vs leathery/shelled eggs on land [1]
- larval stage: aquatic gills vs lungs from birth [1]
(c) max [3]:
- physical traits can look similar due to environmental adaptation / convergent evolution [1]
- DNA base sequence is more accurate / objective [1]
- more similar DNA indicates more recent common ancestor / closer evolutionary relationship [1]
(d) (i) max [2]:
- cell wall [1]
- chloroplasts [1]
- large permanent vacuole [1]
(d) (ii) max [2]:
- prevents cell from bursting (due to osmosis / turgor pressure) [1]
- provides structural support / maintains shape of cell [1]
6 (a) The human gas exchange system is adapted for efficient diffusion of gases.
(i) State the name of the main airway that is kept open by C-shaped rings of cartilage. [1]
(ii) Explain how the structure of alveoli is adapted to maximize the rate of gas exchange. [4]
(b) Goblet cells and ciliated cells work together to defend the gas exchange system against infections.
Describe how these two cell types protect the airways. [3]
(c) Cellular proteins are coded for by genes.
(i) Define the term gene. [2]
(ii) Describe how the sequence of bases in a gene determines the specific structure of a protein. [3]
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解题
(a) (i) Trachea.
(a) (ii) 1. Massive number of alveoli provides a very large total surface area for diffusion.
2. The walls of the alveoli are only one cell thick, which reduces the diffusion distance.
3. A moist lining allows gases to dissolve before diffusing across the membrane.
4. A dense network of blood capillaries maintains a steep concentration gradient by constantly bringing deoxygenated blood and carrying away oxygenated blood.
(b) 1. Goblet cells produce and secrete sticky mucus, which traps inhaled dust particles, bacteria, and pathogens.
2. Ciliated cells have hair-like projections called cilia on their surface.
3. Cilia beat rhythmically to move the mucus (carrying trapped pathogens) upwards away from the lungs toward the throat, where it is swallowed and destroyed by stomach acid.
(c) (i) A gene is a length of DNA that codes for a specific protein.
(c) (ii) 1. The sequence of bases in a gene determines the sequence of bases in mRNA (via transcription).
2. In the cytoplasm, ribosomes read the base sequence in triplets (codons).
3. Each triplet codes for a specific amino acid.
4. This determines the order in which amino acids are joined together to form a polypeptide chain, which then folds into a specific 3-dimensional protein shape.
评分标准
(a) (i) [1]:
- trachea [1]
(a) (ii) max [4]:
- large surface area [1]
- thin wall / one cell thick / short diffusion distance [1]
- moist lining allows gases to dissolve [1]
- dense capillary network / rich blood supply [1]
- ventilation / blood flow maintains concentration gradient [1]
(b) max [3]:
- goblet cells secrete/produce mucus [1]
- mucus traps dust / bacteria / pathogens [1]
- cilia (on ciliated cells) beat / wave [1]
- to sweep mucus away from lungs / to throat [1]
(c) (i) [2]:
- length of DNA [1]
- that codes for a (specific) protein [1]
(c) (ii) max [3]:
- base sequence determines the sequence of amino acids [1]
- three bases (triplet/codon) code for one amino acid [1]
- amino acids are joined in a specific order [1]
- folding determines specific 3D shape / function of the protein [1]
Paper 61 (Alternative to Practical)
The distance moved by the air bubble in 5 minutes was recorded at different wind speeds. The wind speed was controlled by adjusting the speed setting of a nearby electric fan.
Table 1.1 shows the results of this investigation.
**Table 1.1**
| Wind speed / \(m/s\) | Distance moved by the bubble in 5 minutes / \(mm\) |
| :--- | :--- |
| 0.5 | 12 |
| 1.0 | 24 |
| 1.5 | 35 |
| 2.0 | 44 |
| 2.5 | 50 |
| 3.0 | 53 |
(a) Plot a line graph on a grid to show the relationship between wind speed and the distance moved by the bubble in 5 minutes.
(b) Describe the trend shown by the results in Table 1.1.
(c) Calculate the rate of transpiration at a wind speed of 1.5 m/s in mm per minute. Show your working.
(d) Identify:
(i) the independent variable in this investigation.
(ii) two variables that should be kept constant to ensure a valid test.
(e) State one potential source of error when using a bubble potometer and suggest how to minimize its effect.
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解题
(b) As wind speed increases, the distance moved by the bubble increases. The rate of increase slows down (levels off) at higher wind speeds.
(c) Distance at 1.5 m/s is 35 mm.
Time = 5 minutes.
Rate of transpiration = \(35 \div 5 = 7.0\text{ mm/min}\).
(d)(i) Independent variable: Wind speed.
(ii) Variables to control: Temperature, light intensity, humidity, size of shoot, surface area of leaves.
(e) Source of error: Air leaks at the junction between the leafy shoot stem and the potometer. Improvement: Seal the junction securely with petroleum jelly/Vaseline.
评分标准
- Axes correctly labeled with units (x: Wind speed / \(m/s\); y: Distance moved / \(mm\)) [1]
- Linear scale filling at least half of the grid in both directions [1]
- All points plotted accurately [1]
- Points joined with a neat, clean ruled line or smooth curve [1]
(b) [2 marks]
- As wind speed increases, the distance moved by the bubble increases [1]
- The rate of increase slows down / levels off at higher wind speeds [1]
(c) [2 marks]
- Correct working shown: \(35 \div 5\) [1]
- Correct calculation with units: \(7.0\text{ mm/min}\) [1]
(d) [3 marks]
- (i) Wind speed [1]
- (ii) Any two from: temperature, light intensity, humidity, surface area of leaves, duration of measurement [2]
(e) [2 marks]
- Source of error: air leak at the connection / introduction of unwanted air bubbles in potometer capillary tube [1]
- Improvement: seal joints with Vaseline / petroleum jelly, or assemble the apparatus entirely under water [1]
Fig. 2.1 shows a photograph of a sliced broad bean seed showing its internal structure.
[Assume Fig. 2.1 shows a clear internal cross-section of a dicotyledonous seed detailing: testa, cotyledon, radicle, and plumule]
(a) Make a large, clear diagram of the sliced broad bean seed. Label the testa and a cotyledon on your diagram.
(b) The length of the seed in the photograph (line XY) is 45 mm. The actual length of the seed is 15 mm. Calculate the magnification of the seed in Fig. 2.1. Show your working.
(c) Plan an investigation to determine the effect of temperature on the germination of broad bean seeds.
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解题
(b) Magnification = Image length / Actual length = \(45\text{ mm} \div 15\text{ mm} = \times 3\) (or 3).
(c) Investigation plan should clearly describe the independent variable (at least 3 different temperatures, e.g., 5°C, 20°C, 35°C), the dependent variable (counting germinated seeds after a fixed time), at least three control variables (same seed age/source, same volume of water, same light intensity), a step-by-step procedure (e.g. using moist cotton wool in petri dishes, repeating to get reliable average), and safety precautions.
评分标准
- Single clear outlines with no shading [1]
- Drawing larger than the original photograph size [1]
- Plumule and radicle shown clearly in correct proportion [1]
- Correct labels for testa and cotyledon [1]
(b) [2 marks]
- Correct working shown: \(45 \div 15\) [1]
- Correct value: \(\times 3\) or 3 [1]
(c) [7 marks]
- At least three different temperatures tested (e.g., 5°C, 20°C, 35°C) [1]
- Maintain temperature using thermostatically controlled incubators or refrigerator [1]
- Dependent variable defined as counting number of germinated seeds / measuring length of radicle or plumule [1]
- Set time interval for observation (e.g., 5-7 days) [1]
- Control variable 1: same volume of water added to each dish [1]
- Control variable 2: same species / batch / age / size of seeds used [1]
- Control variable 3: same substrate (cotton wool / paper towel) or light conditions [1]
- Set up multiple seeds in each dish (at least 10 seeds) to allow percentage calculation [1]
- Repeat the entire investigation (at least twice) to identify anomalies and calculate averages [1]
- Safety precaution: wash hands after handling seeds / use sterile technique to avoid mold growth [1]
Five potato cylinders of identical length were cut. The initial mass of each cylinder was recorded. Each cylinder was placed into a test-tube containing a different concentration of sucrose solution for 2 hours.
After 2 hours, the cylinders were removed, gently dried, and weighed again.
Table 3.1 shows the results of this investigation.
**Table 3.1**
| Sucrose concentration / \(mol/dm^3\) | Initial mass / \(g\) | Final mass / \(g\) | Percentage change in mass / \(\%\) |
| :--- | :--- | :--- | :--- |
| 0.0 | 4.00 | 4.48 | +12.0 |
| 0.2 | 4.00 | 4.16 | +4.0 |
| 0.4 | 4.00 | 3.84 | -4.0 |
| 0.6 | 4.00 | 3.60 | **[X]** |
| 0.8 | 4.00 | 3.44 | -14.0 |
(a) Calculate the percentage change in mass for the potato cylinder in \(0.6\text{ mol/dm}^3\) sucrose solution, represented by **[X]** in Table 3.1. Show your working.
(b) Plot a line graph of the percentage change in mass against sucrose concentration.
(c) Use your graph to estimate the sucrose concentration that is equal to the concentration of the cell sap inside the potato cells.
(d) State two variables that were kept constant in this investigation.
(e) Explain why the student dried the potato cylinders with a paper towel before weighing them at the end of the experiment.
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解题
(b) Graph should plot percentage change in mass on y-axis (ranging from -15% to +15%) and sucrose concentration on x-axis (0.0 to 0.8 mol/dm³). Points correctly plotted and joined with a clean line of best fit or direct line segments.
(c) The sucrose concentration where percentage change in mass is 0.0% is exactly 0.30 mol/dm³.
(d) Variables kept constant: temperature, duration of immersion (2 hours), dimensions of potato cylinders, same source potato.
(e) To remove excess surface sucrose solution. If not removed, the excess solution would add to the final mass, leading to inaccurate results and an underestimate of mass loss.
评分标准
- Correct working shown: \(\frac{3.60 - 4.00}{4.00} \times 100\) [1]
- Value calculated: 10.0 [1]
- Negative sign or 'decrease' specified [1]
(b) [4 marks]
- Axes labeled with units (x-axis: Sucrose concentration / \(mol/dm^3\); y-axis: Percentage change in mass / \(\%\)) [1]
- Suitable linear scale with both positive and negative values filling at least half of the grid in both directions [1]
- All five points plotted accurately [1]
- Points connected with a straight ruled line or curve of best fit crossing the x-axis [1]
(c) [2 marks]
- Correct concentration read from graph where line crosses y = 0.0% (0.30 ± 0.02 \(mol/dm^3\)) [1]
- Indication on the graph showing how the estimate was obtained [1]
(d) [2 marks]
- Any two from: temperature, immersion time, starting size/dimensions of cylinders, same potato used, same volume of sucrose solution [2]
(e) [3 marks]
- To remove excess sucrose solution from the surface [1]
- Surface solution would add to the final mass [1]
- Would cause an overestimate of mass / lead to inaccurate calculation of percentage change [1]
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