An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V3) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.
Paper 13 (選擇題)
Answer all forty questions. For each question, there are four possible answers, A, B, C, and D. Choose the one correct option.
40 題目 · 40 分
題目 1 · multiple_choice
1 分
An enzyme-catalysed reaction has a \(K_m\) value of \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\) and a \(V_{max}\) of \(50\text{ mmol dm}^{-3}\text{ s}^{-1}\).
The reaction is repeated with the same concentrations of enzyme and substrate, but in the presence of a competitive inhibitor.
Which values of \(K_m\) and \(V_{max}\) could be obtained in the presence of the competitive inhibitor?
A competitive inhibitor competes with the substrate for binding to the active site of the enzyme. This reduces the binding affinity of the enzyme for its substrate, which increases the value of the Michaelis-Menten constant (\(K_m\)). However, because the inhibition can be fully overcome at infinitely high substrate concentrations, the maximum rate of reaction (\(V_{max}\)) remains unchanged. Therefore, in the presence of the competitive inhibitor, the new \(K_m\) must be greater than \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\) (e.g. \(4.0 \times 10^{-4}\text{ mol dm}^{-3}\)), while the \(V_{max}\) must remain at \(50\text{ mmol dm}^{-3}\text{ s}^{-1}\).
評分準則
1 mark for identifying that competitive inhibition increases the \(K_m\) value and leaves the \(V_{max}\) value unchanged, leading to option B.
題目 2 · multiple_choice
1 分
Which row correctly identifies the substances produced during the light-dependent stage of photosynthesis?
A.Produced only by cyclic photophosphorylation: ATP; Produced only by non-cyclic photophosphorylation: reduced NADP and \(O_2\); Produced by both: none
B.Produced only by cyclic photophosphorylation: none; Produced only by non-cyclic photophosphorylation: reduced NADP and \(O_2\); Produced by both: ATP
C.Produced only by cyclic photophosphorylation: ATP and reduced NADP; Produced only by non-cyclic photophosphorylation: \(O_2\); Produced by both: none
D.Produced only by cyclic photophosphorylation: none; Produced only by non-cyclic photophosphorylation: ATP; Produced by both: reduced NADP and \(O_2\)
查看答案詳解收起答案詳解
解題
In cyclic photophosphorylation, excited electrons from Photosystem I (PSI) pass through an electron transport chain and return back to PSI, generating ATP via chemiosmosis. Neither reduced NADP nor oxygen is produced. In non-cyclic photophosphorylation, electrons pass from Photosystem II (PSII) to PSI, generating ATP. The photolysis of water replaces electrons in PSII and produces oxygen (\(O_2\)) and protons. Protons and electrons are used to reduce NADP to reduced NADP. Therefore, reduced NADP and oxygen are produced exclusively by non-cyclic photophosphorylation, ATP is produced by both pathways, and no products are unique to cyclic photophosphorylation.
評分準則
1 mark for identifying the correct distribution of products of cyclic and non-cyclic photophosphorylation.
題目 3 · multiple_choice
1 分
A student calculated Simpson’s Index of Diversity (\(D\)) for the tree species in two different woodlands, woodland P and woodland Q.
Which statement is a correct interpretation of these results?
A.Woodland Q has a more stable ecosystem that is more resilient to environmental changes than woodland P.
B.Woodland P is dominated by a single tree species, whereas woodland Q has a high evenness of species.
C.Two individual trees chosen at random from woodland P are more likely to belong to different species than two trees chosen at random from woodland Q.
D.An increase in the population of one tree species in woodland P is more likely to destabilise the entire community than in woodland Q.
查看答案詳解收起答案詳解
解題
Simpson's Index of Diversity (\(D\)) is the probability that two individuals randomly selected from a sample will belong to different species. A higher value of \(D\) indicates greater species richness and evenness, making the ecosystem more stable and resilient. Because woodland P has a higher value of \(D\) (0.85) than woodland Q (0.35), two trees chosen at random from woodland P are much more likely to belong to different species (85% probability) than two trees chosen at random from woodland Q (35% probability).
評分準則
1 mark for identifying that Simpson's Index of Diversity represents the probability of two randomly selected organisms belonging to different species, and interpreting the higher value in woodland P correctly.
題目 4 · multiple_choice
1 分
Which row correctly describes the pathways of water movement through a plant root and the effect of the Casparian strip?
A.Apoplast pathway: through cytoplasm and plasmodesmata; Symplast pathway: through cell walls and intercellular spaces; Effect of Casparian strip: blocks the apoplast pathway
B.Apoplast pathway: through cell walls and intercellular spaces; Symplast pathway: through cytoplasm and plasmodesmata; Effect of Casparian strip: blocks the symplast pathway
C.Apoplast pathway: through cell walls and intercellular spaces; Symplast pathway: through cytoplasm and plasmodesmata; Effect of Casparian strip: blocks the apoplast pathway
D.Apoplast pathway: through cytoplasm and plasmodesmata; Symplast pathway: through cell walls and intercellular spaces; Effect of Casparian strip: blocks the symplast pathway
查看答案詳解收起答案詳解
解題
The apoplast pathway refers to water and dissolved mineral ions moving through the non-living parts of the cell, specifically the cellulose cell walls and intercellular spaces of the root cortex. The symplast pathway involves water moving from cell to cell via the living cytoplasm and plasmodesmata. The Casparian strip, containing waterproof suberin in the endodermal cell walls, blocks the apoplast pathway, forcing water and dissolved mineral ions to pass through the selectively permeable cell surface membrane into the symplast pathway.
評分準則
1 mark for correctly matching the apoplast pathway to cell walls, the symplast pathway to cytoplasm/plasmodesmata, and identifying that the Casparian strip blocks the apoplast pathway.
題目 5 · multiple_choice
1 分
Which sequence of events correctly describes the action of auxin (IAA) in cell elongation?
A.\(H^+\) ions pumped into cell wall \(\rightarrow\) cell wall pH decreases \(\rightarrow\) expansins activated \(\rightarrow\) hydrogen bonds between cellulose microfibrils broken \(\rightarrow\) water enters cell
B.\(H^+\) ions pumped into cytoplasm \(\rightarrow\) cell wall pH increases \(\rightarrow\) expansins activated \(\rightarrow\) covalent bonds between cellulose microfibrils broken \(\rightarrow\) water leaves cell
C.\(H^+\) ions pumped into cell wall \(\rightarrow\) cell wall pH increases \(\rightarrow\) expansins inactivated \(\rightarrow\) hydrogen bonds between cellulose microfibrils formed \(\rightarrow\) water enters cell
D.\(H^+\) ions pumped into cytoplasm \(\rightarrow\) cell wall pH decreases \(\rightarrow\) expansins inactivated \(\rightarrow\) covalent bonds between cellulose microfibrils broken \(\rightarrow\) water leaves cell
查看答案詳解收起答案詳解
解題
Auxin stimulates proton pumps (\(H^+\)-ATPases) in the cell surface membrane to actively pump hydrogen ions (\(H^+\)) from the cytoplasm into the cell wall. This acidifies the cell wall, decreasing its pH. The lower pH activates expansin proteins, which break the non-covalent hydrogen bonds between cellulose microfibrils, weakening the cell wall. The reduced wall pressure allows water to enter the cell by osmosis down a water potential gradient, causing the cell to expand and elongate.
評分準則
1 mark for identifying the correct sequence of events in auxin-induced cell elongation: \(H^+\) pumped to cell wall, decrease in pH, expansin activation, hydrogen bonds broken, and water entry.
題目 6 · multiple_choice
1 分
In a population of land snails, shell colour is controlled by a single gene with two alleles. The allele for dark shell colour is dominant to the allele for light shell colour.
In this population, 64% of the snails have dark shells.
Assuming the population is in Hardy-Weinberg equilibrium, what is the frequency of the dominant allele?
A.0.36
B.0.40
C.0.60
D.0.80
查看答案詳解收起答案詳解
解題
Let \(p\) be the frequency of the dominant allele and \(q\) be the frequency of the recessive allele. The frequency of snails with dark shells (dominant phenotype) is \(p^2 + 2pq = 0.64\). This means the frequency of snails with light shells (homozygous recessive phenotype, \(q^2\)) is \(1 - 0.64 = 0.36\). Therefore, the frequency of the recessive allele is \(q = \sqrt{0.36} = 0.60\). Since \(p + q = 1\), the frequency of the dominant allele is \(p = 1 - 0.60 = 0.40\).
評分準則
1 mark for calculating the correct frequency of the dominant allele (0.40) by first determining \(q^2 = 0.36\), then \(q = 0.60\), and finally \(p = 0.40\).
題目 7 · multiple_choice
1 分
Which row correctly identifies the presence or absence of tissues in the wall of a trachea, bronchiole, and alveolus?
The trachea wall contains cartilage (C-shaped rings) to prevent collapse, goblet cells to secrete mucus, and smooth muscle to adjust airway diameter. Bronchioles lack cartilage entirely and do not contain goblet cells, but they contain smooth muscle which can contract or relax to regulate airflow to the alveoli. Alveoli are composed of a single layer of squamous epithelium with elastic fibres and completely lack cartilage, goblet cells, and smooth muscle.
評分準則
1 mark for correctly identifying the presence or absence of cartilage, goblet cells, and smooth muscle in the trachea, bronchiole, and alveolus.
題目 8 · multiple_choice
1 分
A cell from a diploid organism has 12 chromosomes in the \(G_1\) phase of the cell cycle.
Which row correctly describes the number of chromosomes, the number of chromatids, and the relative mass of DNA in this cell during prophase of mitosis? (Let the mass of DNA in the \(G_1\) phase be 1.)
A.Number of chromosomes: 12; Number of chromatids: 24; Relative mass of DNA: 2
B.Number of chromosomes: 24; Number of chromatids: 24; Relative mass of DNA: 2
C.Number of chromosomes: 12; Number of chromatids: 12; Relative mass of DNA: 1
D.Number of chromosomes: 24; Number of chromatids: 48; Relative mass of DNA: 4
查看答案詳解收起答案詳解
解題
In the \(G_1\) phase, the cell is diploid and has 12 chromosomes, with each chromosome containing a single double-stranded DNA molecule (relative DNA mass = 1). During the S phase, DNA replication occurs, so each of the 12 chromosomes now consists of 2 identical sister chromatids joined at a single centromere. In prophase of mitosis, the chromosome number is still 12 (as chromosomes are counted by the number of centromeres), but because each chromosome has replicated, there are 24 chromatids. Since replication has doubled the DNA content, the relative mass of DNA is 2.
評分準則
1 mark for correctly identifying that in prophase there are 12 chromosomes, 24 chromatids, and a relative DNA mass of 2.
題目 9 · 選擇題
1 分
An experiment was carried out to study the effects of two inhibitors, X and Y, on an enzyme-catalysed reaction. The Michaelis-Menten constant (\(K_m\)) and maximum velocity (\(V_{max}\)) of the reaction were determined under three conditions: Condition 1 (No inhibitor): \(K_m\) is 2.5 mmol dm\(^{-3}\), \(V_{max}\) is 100 arbitrary units; Condition 2 (With inhibitor X): \(K_m\) is 5.0 mmol dm\(^{-3}\), \(V_{max}\) is 100 arbitrary units; Condition 3 (With inhibitor Y): \(K_m\) is 2.5 mmol dm\(^{-3}\), \(V_{max}\) is 50 arbitrary units. Which statement correctly identifies the types of inhibition shown by X and Y?
A.X is a competitive inhibitor because it increases the affinity of the enzyme for its substrate; Y is a non-competitive inhibitor because it decreases the maximum velocity of the reaction.
B.X is a competitive inhibitor because it decreases the affinity of the enzyme for its substrate; Y is a non-competitive inhibitor because it decreases the maximum velocity of the reaction.
C.X is a non-competitive inhibitor because it increases the Michaelis-Menten constant; Y is a competitive inhibitor because it decreases the maximum velocity of the reaction.
D.X is a non-competitive inhibitor because it does not change the maximum velocity; Y is a competitive inhibitor because it does not change the Michaelis-Menten constant.
查看答案詳解收起答案詳解
解題
For inhibitor X: The \(K_m\) increased from 2.5 to 5.0 mmol dm\(^{-3}\) while \(V_{max}\) remained unchanged at 100. An increase in \(K_m\) reflects a decrease in the affinity of the enzyme for its substrate, which is characteristic of competitive inhibition. For inhibitor Y: The \(V_{max}\) decreased from 100 to 50 arbitrary units while \(K_m\) remained unchanged at 2.5 mmol dm\(^{-3}\). This reduction in maximum rate without affecting substrate affinity is characteristic of non-competitive inhibition.
評分準則
[1 mark] - Correctly identifies X as a competitive inhibitor that decreases affinity (increases \(K_m\)) and Y as a non-competitive inhibitor that reduces \(V_{max}\).
題目 10 · 選擇題
1 分
Which row correctly identifies the features of cyclic photophosphorylation in a chloroplast?
A.Photosystem(s) involved: PS I only; Products formed: ATP only; Fate of excited electrons: Returned to the same reaction centre via electron carriers
B.Photosystem(s) involved: PS I only; Products formed: ATP and reduced NADP; Fate of excited electrons: Used to reduce NADP+ in the stroma
C.Photosystem(s) involved: PS I and PS II; Products formed: ATP only; Fate of excited electrons: Passed to Photosystem I from Photosystem II
D.Photosystem(s) involved: PS II only; Products formed: ATP only; Fate of excited electrons: Returned to the same reaction centre via electron carriers
查看答案詳解收起答案詳解
解題
In cyclic photophosphorylation, only Photosystem I (PS I) is involved. Light energy excites electrons in the reaction centre (P700), which are emitted and passed along an electron transport chain, releasing energy to generate ATP. The electrons are then returned to the same reaction centre. No reduced NADP or oxygen is produced, unlike in non-cyclic photophosphorylation.
評分準則
[1 mark] - Correctly identifies PS I as the only photosystem involved, ATP as the only product, and that the excited electrons are returned to the same reaction centre.
題目 11 · 選擇題
1 分
The number of individuals of four tree species found in two areas of woodland were counted. In Area P, the counts were: Oak = 45, Birch = 5, Beech = 3, Pine = 2 (Total = 55). In Area Q, the counts were: Oak = 12, Birch = 11, Beech = 14, Pine = 13 (Total = 50). Using Simpson's Index of Diversity, \(D = 1 - \sum (n/N)^2\), which statement is correct?
A.Area P has a higher species richness and a higher Simpson's Index of Diversity than Area Q.
B.Area Q has a higher species evenness and a higher Simpson's Index of Diversity than Area P.
C.Area P has a lower species richness and a lower Simpson's Index of Diversity than Area Q.
D.Area Q has a lower species evenness and a higher Simpson's Index of Diversity than Area P.
查看答案詳解收起答案詳解
解題
Both Area P and Area Q contain 4 species, so they have the same species richness. In Area Q, the individuals are more evenly distributed across the species (high species evenness), whereas Area P is dominated by Oak (low species evenness). Calculating Simpson's Index of Diversity: For Area P, \(\sum (n/N)^2 \approx (45/55)^2 + (5/55)^2 + (3/55)^2 + (2/55)^2 = 0.681\), so \(D = 1 - 0.681 = 0.319\). For Area Q, \(\sum (n/N)^2 \approx (12/50)^2 + (11/50)^2 + (14/50)^2 + (13/50)^2 = 0.252\), so \(D = 1 - 0.252 = 0.748\). Thus, Area Q has higher species evenness and a higher Simpson's Index of Diversity.
評分準則
[1 mark] - Recognises that species richness is identical, identifies Area Q as having higher species evenness, and correctly determines that Area Q has a higher Simpson's Index of Diversity.
題目 12 · 選擇題
1 分
Which process is directly responsible for generating the tension (negative pressure) that pulls water up the xylem of an actively transpiring woody plant?
A.Active loading of mineral ions into the root xylem, causing water to enter by osmosis.
B.Evaporation of water from the cell walls of mesophyll cells into the sub-stomatal air spaces.
C.Cohesion between water molecules due to hydrogen bonding as they move up the stem.
D.Adhesion of water molecules to the cellulose in the walls of the xylem vessel elements.
查看答案詳解收起答案詳解
解題
The evaporation of water from the wet cell walls of mesophyll cells into the sub-stomatal air spaces causes water to be drawn from the leaf xylem by osmosis. This creates a tension (negative pressure) at the top of the water column in the xylem vessels. Cohesion and adhesion are physical properties of water that allow this tension to be transmitted down the stem without the column breaking, but they do not generate the pulling force itself.
評分準則
[1 mark] - Identifies evaporation of water from mesophyll cell walls as the generator of negative pressure (tension).
題目 13 · 選擇題
1 分
Which sequence of events correctly describes how auxin (IAA) stimulates cell elongation in a coleoptile during phototropism?
A.Auxin stimulates proton pumps \(\to\) H\(^+\) ions are pumped from the cytoplasm into the cell wall \(\to\) cell wall pH decreases \(\to\) expansins are activated \(\to\) hydrogen bonds between cellulose microfibrils are broken.
B.Auxin stimulates proton pumps \(\to\) H\(^+\) ions are pumped from the cell wall into the cytoplasm \(\to\) cell wall pH increases \(\to\) expansins are activated \(\to\) hydrogen bonds between cellulose microfibrils are broken.
C.Auxin inhibits proton pumps \(\to\) H\(^+\) ions accumulate in the cytoplasm \(\to\) cell wall pH increases \(\to\) expansins are inactivated \(\to\) covalent bonds between cellulose microfibrils are broken.
D.Auxin inhibits proton pumps \(\to\) H\(^+\) ions are pumped from the cytoplasm into the cell wall \(\to\) cell wall pH decreases \(\to\) expansins are activated \(\to\) covalent bonds between cellulose microfibrils are broken.
查看答案詳解收起答案詳解
解題
Under the influence of auxin (IAA), proton pumps in the cell surface membrane actively transport hydrogen ions (H\(^+\)) from the cytoplasm into the cell wall. This lowers the pH of the cell wall (acid growth). The acidic environment activates expansin proteins, which disrupt the hydrogen bonds between cellulose microfibrils. This loosens the cell wall, allowing it to stretch as water enters the vacuole by osmosis and increases turgor pressure.
評分準則
[1 mark] - Identifies correct order of events: proton pump stimulation, H\(^+\) pumping into cell wall, pH decrease, expansin activation, and hydrogen bond cleavage.
題目 14 · 選擇題
1 分
A population of fish in a large lake became separated into two sub-populations when the water level fell, creating two smaller, isolated lakes. Over thousands of years, the two populations evolved different courtship behaviours. When the water level rose again and the two populations merged, they did not interbreed because females did not recognize the courtship displays of males from the other population. Which row correctly classifies the type of speciation and the type of reproductive isolating mechanism that occurred?
A.Type of speciation: Allopatric; Type of reproductive isolating mechanism: Pre-zygotic
B.Type of speciation: Allopatric; Type of reproductive isolating mechanism: Post-zygotic
C.Type of speciation: Sympatric; Type of reproductive isolating mechanism: Pre-zygotic
D.Type of speciation: Sympatric; Type of reproductive isolating mechanism: Post-zygotic
查看答案詳解收起答案詳解
解題
The initial separation was due to a physical geographical barrier (the low water level separating the lake), which leads to allopatric speciation. The reproductive isolating mechanism is behavioral (non-recognition of courtship displays), which prevents mating and zygote formation from occurring in the first place, thus classifying it as a pre-zygotic isolating mechanism.
評分準則
[1 mark] - Identifies geographic separation as allopatric speciation and behavioural barrier as pre-zygotic isolation.
題目 15 · 選擇題
1 分
The tissues present in different parts of the human gas exchange system vary. Which of the following options correctly describes the tissue components present in the specified structure?
A.Trachea: Cartilage is present, Smooth muscle is present, Ciliated epithelium is present, Goblet cells are present
B.Bronchus: Cartilage is absent, Smooth muscle is present, Ciliated epithelium is present, Goblet cells are present
C.Larger Bronchiole: Cartilage is present, Smooth muscle is present, Ciliated epithelium is present, Goblet cells are absent
D.Alveolus: Cartilage is absent, Smooth muscle is present, Ciliated epithelium is absent, Goblet cells are absent
查看答案詳解收起答案詳解
解題
The trachea contains all four specified tissue elements: C-shaped rings of cartilage keep the airway open, smooth muscle (trachealis) allows constriction, ciliated epithelium moves mucus, and goblet cells secrete mucus. Bronchi also contain these tissues, but option B incorrectly states cartilage is absent. Bronchioles lack cartilage (ruling out C), and alveoli consist of squamous epithelium and lack smooth muscle, ciliated epithelium, and goblet cells (ruling out D).
評分準則
[1 mark] - Correctly identifies that the trachea possesses cartilage, smooth muscle, ciliated epithelium, and goblet cells.
題目 16 · 選擇題
1 分
During the complete aerobic respiration of one molecule of glucose, how many molecules of carbon dioxide (\(CO_2\)), reduced NAD, and reduced FAD are produced in total during the link reaction and the Krebs cycle combined?
For each molecule of glucose, two molecules of pyruvate enter the link reaction. The link reaction produces: 2 \(CO_2\) molecules and 2 reduced NAD molecules. The two molecules of acetyl-CoA then enter the Krebs cycle (two turns in total). The Krebs cycle produces: 4 \(CO_2\) molecules, 6 reduced NAD molecules, and 2 reduced FAD molecules. Combining these two stages: Total \(CO_2 = 2 + 4 = 6\); Total reduced NAD = \(2 + 6 = 8\); Total reduced FAD = \(0 + 2 = 2\).
評分準則
[1 mark] - Correctly calculates total molecules produced across both link reaction and Krebs cycle per molecule of glucose.
題目 17 · multiple_choice
1 分
An investigation into enzyme-catalysed reactions measured the effect of an inhibitor on the rate of reaction. In the presence of the inhibitor, the maximum rate of reaction (\(V_{max}\)) remains unchanged, but the Michaelis-Menten constant (\(K_m\)) increases.
Which statement correctly describes the type of inhibition and its effect on the apparent affinity of the enzyme for its substrate?
A.competitive inhibition, which decreases the apparent affinity of the enzyme for its substrate
B.competitive inhibition, which increases the apparent affinity of the enzyme for its substrate
C.non-competitive inhibition, which decreases the apparent affinity of the enzyme for its substrate
D.non-competitive inhibition, which does not affect the apparent affinity of the enzyme for its substrate
查看答案詳解收起答案詳解
解題
Competitive inhibitors bind to the active site, meaning their effects can be overcome by high concentrations of substrate, leaving \(V_{max}\) unchanged. Because more substrate is required to reach half of \(V_{max}\), the Michaelis-Menten constant (\(K_m\)) increases. An increase in \(K_m\) corresponds to a decrease in the apparent affinity of the enzyme for its substrate.
評分準則
Award 1 mark for identifying that the inhibitor is competitive and that an increase in \(K_m\) indicates a decrease in the apparent affinity of the enzyme for its substrate.
題目 18 · multiple_choice
1 分
DCMU is a herbicide that binds to a specific protein in photosystem II (PSII) and blocks the transfer of electrons to the electron transport chain.
Which of the following processes would continue to occur in chloroplasts treated with DCMU?
A.photolysis of water and cyclic photophosphorylation
B.cyclic photophosphorylation only
C.non-cyclic photophosphorylation and the reduction of NADP
D.the reduction of NADP only
查看答案詳解收起答案詳解
解題
DCMU blocks electron flow from photosystem II (PSII). Since PSII cannot pass electrons forward, it cannot remain oxidised and therefore cannot drive the photolysis of water. Consequently, oxygen production and non-cyclic photophosphorylation (which depends on PSII electrons) stop. However, cyclic photophosphorylation only involves photosystem I (PSI) and its associated electron transport chain, recycling electrons to synthesise ATP without requiring PSII or photolysis. Thus, cyclic photophosphorylation continues.
評分準則
Award 1 mark for the correct identification of cyclic photophosphorylation as the only process continuing because it only involves photosystem I.
題目 19 · multiple_choice
1 分
Two areas of woodland, X and Y, were sampled to assess their ground-dwelling beetle diversity. The results were used to calculate Simpson's Index of Diversity (\(D\)) for each area: - Woodland X: \(D = 0.78\) - Woodland Y: \(D = 0.23\)
Which statement is a correct interpretation of these results?
A.A change in environmental conditions is more likely to cause a major disruption to the community in Woodland X than in Woodland Y.
B.Woodland Y has a more stable ecosystem with a greater number of ecological niches available than Woodland X.
C.The ecosystem in Woodland X is more stable, and a change in biotic or abiotic factors is less likely to result in total community collapse than in Woodland Y.
D.Woodland Y is characterized by high species evenness, meaning the individuals are distributed relatively equally among the species.
查看答案詳解收起答案詳解
解題
A high Simpson's Index of Diversity (\(D\)) close to 1 (like Woodland X with \(D = 0.78\)) indicates a highly diverse community with many different species and high evenness. Such ecosystems are typically more stable and less vulnerable to biotic or abiotic changes, meaning a disturbance is unlikely to cause a total collapse. Woodland Y (\(D = 0.23\)) has low diversity, is likely dominated by a single species, and is highly vulnerable to environmental change.
評分準則
Award 1 mark for identifying that higher species diversity (higher \(D\)) makes an ecosystem more stable and less vulnerable to biotic or abiotic disruptions.
題目 20 · multiple_choice
1 分
Which row correctly identifies the mechanisms involved in the active loading of sucrose into the phloem companion cells and its subsequent movement into the sieve tube element?
A.movement of \(H^+\) out of companion cell: active transport; movement of sucrose into companion cell: co-transport with \(H^+\); movement of sucrose into sieve tube: diffusion through plasmodesmata
B.movement of \(H^+\) out of companion cell: facilitated diffusion; movement of sucrose into companion cell: active transport; movement of sucrose into sieve tube: active transport
C.movement of \(H^+\) out of companion cell: active transport; movement of sucrose into companion cell: facilitated diffusion; movement of sucrose into sieve tube: active transport
D.movement of \(H^+\) out of companion cell: osmosis; movement of sucrose into companion cell: co-transport with \(H^+\); movement of sucrose into sieve tube: diffusion through plasmodesmata
查看答案詳解收起答案詳解
解題
Active loading begins with companion cells actively pumping hydrogen ions (\(H^+\)) out of their cytoplasm into the cell wall (apoplast) using ATP. This creates a proton gradient. \(H^+\) ions then diffuse back down their gradient into the companion cell via co-transporter proteins, bringing sucrose molecules with them against their concentration gradient. Finally, sucrose moves passively from the companion cell into the sieve tube element by diffusion through connecting plasmodesmata.
評分準則
Award 1 mark for the correct combination showing active transport of protons, co-transport of sucrose, and passive diffusion of sucrose through plasmodesmata.
題目 21 · multiple_choice
1 分
During the germination of barley seeds, gibberellin stimulates the synthesis of \(\alpha\)-amylase. An experiment is set up where barley seeds with their embryos removed are incubated in different test solutions: - Group 1: incubated in distilled water only - Group 2: incubated in gibberellin solution - Group 3: incubated in gibberellin solution containing an RNA polymerase inhibitor - Group 4: incubated in gibberellin solution containing a eukaryotic translation inhibitor
In which of the groups will \(\alpha\)-amylase be successfully produced by the aleurone layer?
A.Group 2 only
B.Group 2 and Group 3 only
C.Group 2, Group 3 and Group 4
D.Group 1 and Group 2 only
查看答案詳解收起答案詳解
解題
Gibberellin triggers the germination of barley seeds by binding to receptors in the aleurone layer and stimulating the transcription of the gene encoding \(\alpha\)-amylase, followed by translation of the mRNA. Group 1 has no embryo to produce natural gibberellin, and water alone does not stimulate synthesis. Group 3 contains an RNA polymerase inhibitor which blocks transcription. Group 4 contains a translation inhibitor which blocks protein synthesis. Therefore, only Group 2 can successfully produce the enzyme.
評分準則
Award 1 mark for identifying that Group 2 is the only group that allows both transcription and translation to occur in response to gibberellin.
題目 22 · multiple_choice
1 分
A species of wild grass grows on soil contaminated with high levels of copper from a nearby industrial site. Most plants of this species cannot tolerate high copper concentrations. However, a few individuals possess a mutant allele that allows them to tolerate copper. Over several generations, the proportion of copper-tolerant plants in the population on the contaminated soil increases.
Which statements about this population are correct? 1. The copper contamination acts as a selective agent. 2. This is an example of stabilizing selection. 3. The copper-tolerance allele arose in response to the plants being exposed to copper. 4. Plants with the copper-tolerance allele have a higher selective advantage on the contaminated soil.
A.1, 2 and 3
B.1 and 4 only
C.2 and 4 only
D.1, 3 and 4
查看答案詳解收起答案詳解
解題
Statement 1 is correct because the toxic copper in the soil represents an environmental pressure selecting for tolerant individuals. Statement 4 is correct because tolerant plants survive, reproduce, and pass on the beneficial allele. Statement 2 is incorrect because this selection shifts the population towards one phenotypic extreme (copper tolerance), which is directional selection. Statement 3 is incorrect because mutations occur randomly and do not arise as a direct response to exposure to an environmental factor.
評分準則
Award 1 mark for identifying that only statements 1 and 4 are correct.
題目 23 · multiple_choice
1 分
Which statement correctly describes the presence or absence of tissue components in the walls of mammalian bronchioles and alveoli?
A.Bronchioles have elastic fibres and smooth muscle, but lack cartilage; alveoli have elastic fibres, but lack smooth muscle and cartilage.
B.Bronchioles have cartilage and smooth muscle, but lack elastic fibres; alveoli have elastic fibres and smooth muscle, but lack cartilage.
C.Bronchioles have elastic fibres, but lack smooth muscle and cartilage; alveoli have elastic fibres and smooth muscle, but lack cartilage.
D.Bronchioles have cartilage, smooth muscle and elastic fibres; alveoli have elastic fibres, but lack smooth muscle and cartilage.
查看答案詳解收起答案詳解
解題
Bronchioles lack cartilage but have smooth muscle and elastic fibres in their walls to control the diameter of the airway. Alveolar walls consist of simple squamous epithelium supported by elastic fibres (to allow expansion and elastic recoil during ventilation) but completely lack cartilage and smooth muscle.
評分準則
Award 1 mark for identifying that bronchioles have elastic fibres and smooth muscle but lack cartilage, whereas alveoli have elastic fibres but lack smooth muscle and cartilage.
題目 24 · multiple_choice
1 分
A diploid eukaryotic cell from an organism with a chromosome number of \(2n = 12\) undergoes mitosis.
How many chromosomes and how many DNA molecules are present in this single cell during metaphase and during anaphase?
A.metaphase: 12 chromosomes and 24 DNA molecules; anaphase: 24 chromosomes and 24 DNA molecules
B.metaphase: 12 chromosomes and 12 DNA molecules; anaphase: 12 chromosomes and 24 DNA molecules
C.metaphase: 12 chromosomes and 24 DNA molecules; anaphase: 12 chromosomes and 12 DNA molecules
D.metaphase: 24 chromosomes and 24 DNA molecules; anaphase: 24 chromosomes and 12 DNA molecules
查看答案詳解收起答案詳解
解題
At metaphase, there are 12 chromosomes lined up at the spindle equator. Since DNA replication occurred during S phase, each chromosome consists of two sister chromatids, resulting in 24 DNA molecules. During anaphase, the centromeres split and sister chromatids separate to opposite poles. Each chromatid is now considered an individual chromosome, making the total chromosome number in the cell 24. No cytokinesis has occurred yet, so there are still 24 DNA molecules in the cell.
評分準則
Award 1 mark for identifying that metaphase has 12 chromosomes and 24 DNA molecules, and anaphase has 24 chromosomes and 24 DNA molecules.
題目 25 · 選擇題
1 分
An enzyme-controlled reaction is carried out with and without different inhibitors. The rate of reaction at different substrate concentrations is measured.
In the presence of inhibitor X, the Michaelis-Menten constant (\(K_m\)) is increased but the maximum velocity (\(V_{\max}\)) remains unchanged.
In the presence of inhibitor Y, the \(V_{\max}\) is decreased but the \(K_m\) remains unchanged.
Which of the following statements about these inhibitors are correct?
1. Inhibitor X has a molecular structure similar to the substrate and binds to the active site. 2. Increasing the substrate concentration can overcome the inhibitory effect of Y. 3. Inhibitor Y binds to an allosteric site on the enzyme, changing the shape of the active site. 4. Inhibitor X decreases the affinity of the enzyme for its substrate.
A.1, 2 and 3
B.1, 3 and 4
C.2 and 4 only
D.1 and 3 only
查看答案詳解收起答案詳解
解題
Inhibitor X increases \(K_m\) without changing \(V_{\max}\), which is characteristic of a competitive inhibitor. Competitive inhibitors have a structure similar to the substrate, bind to the active site (statement 1), and decrease the apparent affinity of the enzyme for its substrate (statement 4). Inhibitor Y decreases \(V_{\max}\) without changing \(K_m\), which is characteristic of a non-competitive inhibitor. Non-competitive inhibitors bind to an allosteric site, altering the active site's shape (statement 3), and their inhibitory effect cannot be overcome by increasing substrate concentration (making statement 2 incorrect). Thus, statements 1, 3, and 4 are correct.
評分準則
Award 1 mark for the correct option (B). No partial marks are awarded for multiple-choice questions.
題目 26 · 選擇題
1 分
During photosynthesis, light energy is used to synthesise ATP and reduced NADP through photophosphorylation.
Which row in the table correctly describes the effects of an inhibitor that blocks electron transfer from the primary electron acceptor of Photosystem II (PSII) to the electron transport chain linking PSII and PSI?
If electron transfer from the primary electron acceptor of Photosystem II (PSII) is blocked, the non-cyclic pathway of photophosphorylation is interrupted. As a result, PSII cannot pass its excited electrons down the transport chain, stopping the photolysis of water (so oxygen production stops). Since electrons cannot flow from PSII to PSI, the production of reduced NADP also stops. However, cyclic photophosphorylation only involves Photosystem I (PSI) and its associated electron transport loop to generate ATP, so cyclic ATP production continues.
評分準則
Award 1 mark for the correct option (A). No partial marks are awarded for multiple-choice questions.
題目 27 · 選擇題
1 分
The table shows the number of individuals of three different tree species found in two equal-sized areas of woodland, Area X and Area Y.
| Species | Number of individuals in Area X | Number of individuals in Area Y | |---|---|---| | Species A | 10 | 27 | | Species B | 10 | 2 | | Species C | 10 | 1 |
Which statement about the biodiversity of these two areas is correct?
A.Area X has a lower species richness and a higher Simpson's Index of Diversity than Area Y.
B.Area Y has a higher species richness and a lower Simpson's Index of Diversity than Area X.
C.Both areas have the same species richness, but Area X has a higher Simpson's Index of Diversity than Area Y.
D.Both areas have the same species richness, but Area Y has a higher Simpson's Index of Diversity than Area X.
查看答案詳解收起答案詳解
解題
Species richness is simply the number of different species present in an area. Both Area X and Area Y contain exactly three species (A, B, and C), so their species richness is the same. Simpson's Index of Diversity (\(D\)) takes into account both species richness and species evenness (how evenly distributed the individuals are among the species). Area X has high evenness (10 of each), whereas Area Y is dominated by one species (27 of Species A). A higher species evenness results in a higher Simpson's Index of Diversity. Therefore, Area X has a higher Simpson's Index of Diversity than Area Y.
評分準則
Award 1 mark for the correct option (C). No partial marks are awarded for multiple-choice questions.
題目 28 · 選擇題
1 分
Which processes are involved in the movement of hydrogen ions (\(H^+\)) and sucrose molecules during the active loading of sucrose into phloem sieve tube elements at a source?
A.Movement of \(H^+\) out of companion cell: Active transport; Movement of sucrose into companion cell: Co-transport; Movement of sucrose into sieve tube element: Diffusion through plasmodesmata
B.Movement of \(H^+\) out of companion cell: Active transport; Movement of sucrose into companion cell: Facilitated diffusion; Movement of sucrose into sieve tube element: Active transport
C.Movement of \(H^+\) out of companion cell: Facilitated diffusion; Movement of sucrose into companion cell: Co-transport; Movement of sucrose into sieve tube element: Active transport
D.Movement of \(H^+\) out of companion cell: Facilitated diffusion; Movement of sucrose into companion cell: Facilitated diffusion; Movement of sucrose into sieve tube element: Diffusion through plasmodesmata
查看答案詳解收起答案詳解
解題
Active loading of sucrose involves proton pumps actively transporting \(H^+\) ions out of the companion cell into the cell wall apoplast using ATP. This creates an electrochemical gradient. \(H^+\) ions then diffuse back into the companion cell down their gradient via a co-transporter protein, which simultaneously transports sucrose molecules into the companion cell against their concentration gradient (co-transport). Once inside the companion cell, sucrose molecules diffuse into the adjacent sieve tube elements through plasmodesmata.
評分準則
Award 1 mark for the correct option (A). No partial marks are awarded for multiple-choice questions.
題目 29 · 選擇題
1 分
During the germination of barley seeds, gibberellin (GA) stimulates the synthesis of amylase in the aleurone layer.
What is the correct sequence of events that occurs after gibberellin is released by the embryo?
1. Gibberellin diffuses into the cells of the aleurone layer. 2. Transcription of the gene for amylase occurs, producing mRNA. 3. Amylase hydrolyses starch in the endosperm to maltose. 4. Gibberellin triggers the breakdown of DELLA proteins. 5. Amylase is synthesised and secreted into the endosperm.
Following its release by the embryo, gibberellin (GA) diffuses into the aleurone layer (1). Inside the aleurone cells, GA binds to its receptor, which initiates a signaling cascade that triggers the breakdown of DELLA proteins (4). DELLA proteins normally act as repressors of transcription factors; once they are broken down, transcription of the amylase gene can proceed (2). Amylase is then translated, processed, and secreted into the endosperm (5), where it hydrolyses stored starch into maltose (3).
評分準則
Award 1 mark for the correct option (B). No partial marks are awarded for multiple-choice questions.
題目 30 · 選擇題
1 分
In a large, randomly mating population of wildflowers, the allele for red flowers (\(R\)) is completely dominant to the allele for white flowers (\(r\)).
A survey of the population shows that 9% of the plants have white flowers.
Assuming the population is in Hardy-Weinberg equilibrium, what is the percentage of plants in the population that are heterozygous for flower colour?
A.21%
B.30%
C.42%
D.49%
查看答案詳解收起答案詳解
解題
Using the Hardy-Weinberg equations: \(p^2 + 2pq + q^2 = 1\) and \(p + q = 1\)
Let \(q\) be the frequency of the recessive allele (\(r\)). Plants with white flowers are homozygous recessive (\(rr\)), so: \(q^2 = 0.09 \implies q = \sqrt{0.09} = 0.3\)
Since \(p + q = 1\), the frequency of the dominant allele (\(R\)) is: \(p = 1 - 0.3 = 0.7\)
The frequency of heterozygous plants (\(Rr\)) is given by \(2pq\): \(2pq = 2 \times 0.7 \times 0.3 = 0.42\), which is 42%.
評分準則
Award 1 mark for the correct option (C). No partial marks are awarded for multiple-choice questions.
題目 31 · 選擇題
1 分
Which row in the table correctly describes the presence (\(\checkmark\)) or absence (\(\times\)) of tissues in a human bronchiole and an alveolus?
Bronchioles do not contain cartilage (\(\times\)) but do contain smooth muscle (\(\checkmark\)) to control the diameter of the airway. Alveoli do not have ciliated epithelium (\(\times\))—they are lined with simple squamous epithelium to allow rapid gas exchange—but they do contain elastic fibres (\(\checkmark\)) which stretch during inhalation and recoil to push air out during exhalation. Therefore, Row A is correct.
評分準則
Award 1 mark for the correct option (A). No partial marks are awarded for multiple-choice questions.
題目 32 · 選擇題
1 分
A diploid eukaryotic cell has a chromosome number of \(2n = 12\).
Which row in the table correctly shows the number of chromosomes and the number of DNA molecules present in this cell during G2 phase of interphase and during anaphase of mitosis?
During the G2 phase of interphase, DNA replication has already occurred in the S phase. Each of the 12 chromosomes consists of 2 sister chromatids held together at a centromere, so there are still 12 chromosomes, but there are 24 DNA molecules. During anaphase of mitosis, the centromeres divide and the sister chromatids separate to become individual chromosomes. This temporarily doubles the chromosome number inside the single cell to 24. Since no DNA synthesis or degradation occurs during mitosis, the number of DNA molecules in the cell remains 24. Thus, Row A is correct.
評分準則
Award 1 mark for the correct option (A). No partial marks are awarded for multiple-choice questions.
題目 33 · multiple_choice
1 分
An experiment was carried out to investigate the effect of a competitive inhibitor on an enzyme-catalyzed reaction. Which row correctly describes the effect of increasing the concentration of the competitive inhibitor on the maximum rate of reaction (\(V_{max}\)) and the Michaelis-Menten constant (\(K_m\))?
A.Decrease in \(V_{max}\) and increase in \(K_m\)
B.No change in \(V_{max}\) and increase in \(K_m\)
C.Decrease in \(V_{max}\) and no change in \(K_m\)
D.No change in \(V_{max}\) and decrease in \(K_m\)
查看答案詳解收起答案詳解
解題
Competitive inhibitors bind reversibly to the active site of the enzyme, competing directly with the substrate. At very high substrate concentrations, the substrate can outcompete the inhibitor, so the maximum rate of reaction (\(V_{max}\)) remains unchanged. However, a higher concentration of substrate is required to reach half of the \(V_{max}\), meaning that the Michaelis-Menten constant (\(K_m\)) increases.
評分準則
1 mark for identifying that \(V_{max}\) remains constant and \(K_m\) increases.
題目 34 · multiple_choice
1 分
Which of the following events occurs during cyclic photophosphorylation in the chloroplasts of a healthy leaf cell?
A.Photolysis of water to release oxygen gas
B.Reduction of NADP to NADPH
C.Pumping of protons into the thylakoid lumen to generate a proton gradient
D.Utilization of both Photosystem I and Photosystem II
查看答案詳解收起答案詳解
解題
During cyclic photophosphorylation, only Photosystem I is involved. Photoexcited electrons pass along an electron transport chain, releasing energy used to pump protons from the stroma into the thylakoid lumen, establishing a proton gradient used for ATP synthesis. Because Photosystem II is not involved, there is no photolysis of water, no oxygen is produced, and NADP is not reduced.
評分準則
1 mark for identifying proton pumping into the thylakoid lumen as the correct event.
題目 35 · multiple_choice
1 分
Which statement correctly describes the pathway taken by water and dissolved mineral ions as they move through the root tissues of a plant?
A.Water moving via the apoplast pathway crosses cell surface membranes by osmosis.
B.The Casparian strip forces water in the symplast pathway to enter the apoplast pathway.
C.Water moving via the symplast pathway passes through plasmodesmata between adjacent cells.
D.Solutes in the apoplast pathway require active transport proteins to cross the endodermal cell walls.
查看答案詳解收起答案詳解
解題
In the symplast pathway, water moves through the cytoplasm of cells, which are interconnected by plasmodesmata. The apoplast pathway involves water moving through the cell walls without crossing membranes until it reaches the endodermis, where the Casparian strip (composed of suberin) blocks the apoplastic movement, forcing water and solutes to cross the cell surface membrane into the symplast pathway.
評分準則
1 mark for identifying the correct movement of water through the symplast pathway.
題目 36 · multiple_choice
1 分
A student calculated Simpson's Index of Diversity (\(D\)) for two different fields using the formula: \(D = 1 - \sum (\frac{n}{N})^2\). Field X had a value of \(D = 0.85\) and Field Y had a value of \(D = 0.23\). Which statement is a correct interpretation of these results?
A.Field X has lower species richness and lower species evenness than Field Y.
B.A small change in environmental conditions is more likely to cause a major change in the community of Field X than Field Y.
C.An organism chosen at random from Field Y is highly likely to belong to the same species as another organism chosen at random from Field Y.
D.Field X is dominated by a single highly abundant species compared to Field Y.
查看答案詳解收起答案詳解
解題
A value of \(D\) close to 0 indicates low biodiversity, which typically means the community is dominated by one or two species. Thus, two randomly chosen individuals from Field Y (where \(D = 0.23\)) are highly likely to belong to the same dominant species. Field X has high diversity, meaning high species richness and evenness, making it more stable to environmental changes.
評分準則
1 mark for identifying the correct interpretation of the lower Simpson's Index of Diversity value.
題目 37 · multiple_choice
1 分
How does auxin (IAA) stimulate the elongation of plant cells during phototropism?
A.Auxin binds to cell surface receptors, stimulating protons (\(H^+\)) to be actively pumped into the cell wall, which lowers the pH and activates expansin proteins to loosen cellulose microfibrils.
B.Auxin diffuses down the illuminated side of the shoot, causing cells on that side to divide rapidly by mitosis.
C.Auxin activates the active transport of calcium ions into the vacuole, lowering water potential and causing water to enter by osmosis.
D.Auxin directly breaks the hydrogen bonds between cellulose microfibrils without the involvement of any enzymes or proteins.
查看答案詳解收起答案詳解
解題
Auxin stimulates proton pumps in the cell surface membrane, pumping protons into the cell wall. The resulting decrease in pH activates pH-dependent enzymes called expansins. Expansins break the non-covalent bonds between cellulose microfibrils, loosening the wall structure so that turgor pressure can stretch the cell.
評分準則
1 mark for the correct description of the acid growth hypothesis involving proton pumping and expansins.
題目 38 · multiple_choice
1 分
Some bacteria carry a mutation in the gene encoding the enzyme RNA polymerase, which makes them resistant to the antibiotic rifampicin. In the presence of rifampicin, what describes the selection pressure and the change in allele frequency in this bacterial population?
A.The selection pressure is the mutation; the frequency of the resistant allele decreases.
B.The selection pressure is the mutation; the frequency of the resistant allele increases.
C.The selection pressure is the antibiotic rifampicin; the frequency of the resistant allele decreases.
D.The selection pressure is the antibiotic rifampicin; the frequency of the resistant allele increases.
查看答案詳解收起答案詳解
解題
The selection pressure is the environmental factor that determines survival, which in this case is the antibiotic rifampicin. Bacteria carrying the resistant allele have a selective advantage, survive, and reproduce, passing this allele to their offspring, thereby increasing the allele frequency in the population.
評分準則
1 mark for correctly identifying the selection pressure as the antibiotic and the change in allele frequency as an increase.
題目 39 · multiple_choice
1 分
Which tissues are present in the wall of a bronchiole?
A.Cartilage, smooth muscle, elastic fibres, and ciliated epithelium
B.Smooth muscle, elastic fibres, and ciliated epithelium, but no cartilage
C.Cartilage, elastic fibres, and goblet cells, but no smooth muscle
D.Cartilage, smooth muscle, and goblet cells, but no elastic fibres
查看答案詳解收起答案詳解
解題
Bronchioles lack cartilage, which is found in the trachea and bronchi. They do contain smooth muscle to regulate airflow, elastic fibres to allow expansion and recoil, and are lined with ciliated epithelium.
評分準則
1 mark for identifying the correct combination of tissues present in bronchioles, specifically the absence of cartilage.
題目 40 · multiple_choice
1 分
A student prepared a root tip squash of Allium cepa and counted the number of cells in different stages of the cell cycle. The results were: Interphase = 422 cells, Prophase = 34 cells, Metaphase = 12 cells, Anaphase = 8 cells, Telophase = 14 cells. What is the mitotic index of this tissue?
A.13.9%
B.11.4%
C.12.0%
D.15.6%
查看答案詳解收起答案詳解
解題
The mitotic index is the percentage of cells undergoing mitosis. Number of cells in mitosis = 34 (Prophase) + 12 (Metaphase) + 8 (Anaphase) + 14 (Telophase) = 68. Total number of cells = 422 (Interphase) + 68 (Mitosis) = 490. Mitotic Index = (68 / 490) * 100% = 13.88%, which rounds to 13.9%.
評分準則
1 mark for calculating the correct mitotic index by dividing the mitotic cells by total cells and converting to a percentage.
Paper 23 (AS Level Structured)
Answer all six structured questions. Show all working and use appropriate units.
6 題目 · 60 分
題目 1 · structured
10 分
Amylase is an enzyme that catalyses the hydrolysis of starch to maltose.
(a) Define the term *activation energy*. [2]
(b) Explain, in terms of collision theory and kinetic energy, why the rate of an amylase-catalysed reaction decreases when the temperature is raised significantly above its optimum temperature. [4]
(c) A student investigated the effect of pH on the activity of amylase. State how the student could maintain a constant temperature throughout this investigation, and explain why temperature must be controlled. [4]
查看答案詳解收起答案詳解
解題
(a) Activation energy is the minimum energy required to start a chemical reaction, which allows reactant molecules to reach their transition state and break chemical bonds.
(b) As temperature is increased significantly above the optimum, the kinetic energy of the amylase molecules increases, causing them to vibrate excessively. This breaks the weak hydrogen bonds and ionic bonds maintaining the tertiary structure of the protein. Consequently, the specific shape of the active site changes and is no longer complementary to the starch substrate. The enzyme is denatured, preventing the formation of successful enzyme-substrate (E-S) complexes.
(c) The temperature can be kept constant using a thermostatically controlled water bath. It must be controlled because changes in temperature affect the kinetic energy of the molecules and alter the rate of reaction. If not controlled, temperature would act as an uncontrolled variable, making the results invalid.
評分準則
(a) [Max 2 marks] - Minimum energy required to start a chemical reaction [1] - Allows reactant molecules to reach transition state / break bonds [1]
(b) [Max 4 marks] - Increased kinetic energy leads to increased vibrations of the enzyme molecule [1] - Breaking of hydrogen bonds / ionic bonds / hydrophobic interactions [1] (Reject: peptide or covalent bonds) - Loss of tertiary structure of the enzyme / denaturation [1] - Active site changes shape and is no longer complementary to the substrate [1] - No enzyme-substrate (E-S) complexes can form [1]
(c) [Max 4 marks] - Use of a thermostatically controlled water bath [1] - Use of a thermometer to monitor/record temperature [1] - Temperature affects kinetic energy of reactant/enzyme molecules [1] - If temperature varies, it alters the rate of reaction independently of pH / ensures validity of results [1]
題目 2 · structured
10 分
(a) Define *water potential* and state the water potential value of pure water at atmospheric pressure and standard temperature. [2]
(b) Plant cells were placed in a concentrated (hypertonic) sucrose solution. Describe and explain the appearance of the cells after 30 minutes. [4]
(c) Explain how facilitated diffusion differs from active transport in terms of the membrane proteins involved and energy requirements. [4]
查看答案詳解收起答案詳解
解題
(a) Water potential is the measure of the tendency of water molecules to move from one region to another by osmosis (or a measure of the free energy of water molecules). The water potential of pure water under standard conditions is defined as \(0\text{ kPa}\).
(b) The plant cells will become plasmolysed. The concentrated sucrose solution has a lower water potential than the cell cytoplasm and vacuole. Water moves out of the vacuole and cytoplasm down a water potential gradient by osmosis, passing across the selectively permeable cell membrane. As a result, the volume of the vacuole decreases, and the protoplast shrinks and pulls away from the cell wall, leaving a space filled with sucrose solution.
(c) Facilitated diffusion is a passive process that does not require metabolic energy (ATP) because molecules move down their concentration gradient using kinetic energy. In contrast, active transport is an active process that requires ATP to move molecules against their concentration gradient. Facilitated diffusion uses both channel proteins (which can be gated) and carrier proteins, whereas active transport relies solely on specific carrier proteins (often called pumps) that undergo conformational changes powered by ATP hydrolysis.
評分準則
(a) [Max 2 marks] - Measure of the tendency of water molecules to move / free energy of water [1] - Value is \(0\text{ kPa}\) / zero [1]
(b) [Max 4 marks] - Cells become plasmolysed / flaccid [1] - Sucrose solution has lower water potential / cell has higher water potential [1] - Water moves out of the cell by osmosis [1] - Down a water potential gradient / from high to low water potential [1] - Protoplast shrinks / pulls away from the cell wall [1]
(c) [Max 4 marks] - Facilitated diffusion is passive / does not require ATP AND active transport is active / requires ATP [1] - Facilitated diffusion goes down a concentration gradient AND active transport goes against a concentration gradient [1] - Facilitated diffusion uses both channel and carrier proteins [1] - Active transport uses carrier proteins / pumps only [1] - Active transport carrier proteins undergo conformational changes powered by ATP hydrolysis [1]
題目 3 · structured
10 分
(a) Contrast the *apoplast* pathway and the *symplast* pathway of water movement through the root cortex. [3]
(b) Describe how the structure of a xylem vessel element is adapted to its function of transporting water. [4]
(c) Xerophytic plants have anatomical adaptations to reduce water loss. State three such adaptations and explain how each reduces the rate of transpiration. [3]
查看答案詳解收起答案詳解
解題
(a) The apoplast pathway is the movement of water through the non-living cell walls and intercellular spaces of the root cortex, whereas the symplast pathway involves water moving through the living cytoplasm and plasmodesmata. Water moves faster and with less resistance in the apoplast pathway than in the symplast pathway, and does not have to cross plasma membranes until it reaches the Casparian strip.
(b) Xylem vessel elements are dead cells aligned end-to-end with no cross-walls (end walls), forming a continuous, hollow tube that minimizes resistance to water flow. The absence of cytoplasm and organelles also ensures an unobstructed lumen. Their cell walls are heavily thickened with lignin, which provides high mechanical strength to prevent the vessel from collapsing inward under the extreme negative pressure (tension) generated by transpiration, and is waterproof, preventing lateral water leakage.
(c) 1. Rolled leaves: traps a layer of moist, humid air next to the stomata, which reduces the water potential gradient between the inside and outside of the leaf. 2. Sunken stomata (in pits): traps moist air in the pits, reducing the diffusion gradient for water vapor. 3. Thick waxy cuticle: increases the diffusion path length and acts as a waterproof barrier to minimize cuticular water loss.
評分準則
(a) [Max 3 marks] - Apoplast involves cell walls / intercellular spaces AND symplast involves cytoplasm / plasmodesmata [1] - Apoplast is non-living AND symplast is living [1] - Apoplast is faster / has less resistance to flow than symplast [1] - Water in apoplast does not cross plasma membranes (until endodermis) [1]
(b) [Max 4 marks] - End walls broken down / continuous hollow tubes to allow uninterrupted columns of water [1] - No cytoplasm / no organelles to minimize resistance to water flow [1] - Cell walls strengthened with lignin to prevent collapse under tension [1] - Lignin is waterproof to prevent water escaping [1] - Pits in walls allow lateral movement of water [1]
(c) [Max 3 marks - each point must pair an adaptation with its correct explanation] - Rolled leaves AND traps humid air to reduce water potential gradient [1] - Sunken stomata / stomata in pits AND traps moist air to reduce diffusion gradient [1] - Thick waxy cuticle AND increases diffusion distance / waterproof barrier to reduce cuticular transpiration [1] - Reduced leaves / spines AND reduces surface area for transpiration [1] - Hairs on leaves AND traps moist air / reduces air flow over stomata [1]
題目 4 · structured
10 分
(a) Explain how the alveoli in human lungs are adapted to maximize the rate of gas exchange. [3]
(b) Describe the roles of goblet cells and ciliated epithelial cells in protecting the gas exchange system from pathogens. [4]
(c) Explain how tar in tobacco smoke damages the structure of the gas exchange system and contributes to chronic bronchitis. [3]
查看答案詳解收起答案詳解
解題
(a) Alveoli maximize gas exchange through several adaptations: they provide an extremely large total surface area for diffusion; they have thin walls consisting of a single layer of squamous epithelial cells, which minimizes the diffusion distance; and they are closely associated with an extensive, dense network of capillaries, which, along with ventilation, maintains a steep concentration gradient for oxygen and carbon dioxide.
(b) Goblet cells synthesize and secrete sticky mucus onto the surface of the airway epithelium. This mucus acts to trap inhaled pathogens, dust, and other foreign particles. Ciliated epithelial cells possess hair-like projections called cilia on their apical membranes. These cilia beat in a coordinated, rhythmic fashion to sweep the trapped pathogens and mucus upward away from the lungs toward the throat, where it can be swallowed and destroyed by stomach acid or expectorated.
(c) Tar in tobacco smoke stimulates the goblet cells and mucous glands to enlarge and secrete excess mucus. Simultaneously, tar paralyzes and eventually destroys the cilia on the ciliated epithelial cells. This combination prevents the removal of mucus, which accumulates in the bronchi and bronchioles. This trapped, stagnant mucus obstructs the airways and becomes a breeding ground for bacteria, leading to persistent inflammation and the airway infections characteristic of chronic bronchitis.
評分準則
(a) [Max 3 marks] - Large total surface area for rapid diffusion of gases [1] - Wall is one cell thick / squamous epithelium for short diffusion distance [1] - Dense network of capillaries maintains a steep concentration gradient [1] - Surfactant layer prevents alveolar collapse [1]
(b) [Max 4 marks] - Goblet cells produce / secrete mucus [1] - Mucus is sticky and traps bacteria / pathogens / dust [1] - Ciliated cells have cilia [1] - Cilia beat in a coordinated wave-like manner [1] - Sweeps trapped mucus and pathogens upward / away from the lungs to the throat [1]
(c) [Max 3 marks] - Tar stimulates goblet cells / mucous glands to secrete excess mucus [1] - Tar paralyzes / damages / destroys cilia [1] - Mucus accumulates / is not swept away, leading to blocked airways / bacterial infection / inflammation of bronchi [1]
題目 5 · structured
10 分
(a) Mitosis is part of the cell cycle. State the stage of mitosis in which each of the following events occurs:
(i) Chromosomes align at the equator of the spindle. [1]
(ii) Sister chromatids separate and are pulled to opposite poles. [1]
(iii) The nuclear envelope re-forms around two sets of chromosomes. [1]
(b) Explain the importance of mitosis in living organisms. [4]
(c) Taxol is a drug used in cancer chemotherapy that prevents the breakdown of spindle fibres, keeping them locked in place. Suggest and explain how Taxol affects the process of mitosis and cell division. [3]
查看答案詳解收起答案詳解
解題
(a) (i) Metaphase (ii) Anaphase (iii) Telophase
(b) Mitosis is essential for producing genetically identical daughter cells with the same diploid number of chromosomes as the parent cell. This is vital for: 1. Growth: increasing the number of cells in multicellular organisms. 2. Repair of tissues / replacement of cells: replacing damaged or worn-out cells with exact copies (Reject: cell repair). 3. Asexual reproduction: producing genetically identical offspring in unicellular eukaryotes and some multicellular organisms. 4. Immune response: allowing the clonal expansion of specific B- and T-lymphocytes upon activation by antigens.
(c) During normal anaphase, spindle fibres must depolymerise (shorten) to pull sister chromatids to opposite poles of the cell. If Taxol prevents the breakdown of spindle fibres, chromatids cannot be separated and moved apart. This arrests mitosis at metaphase/anaphase, preventing the cell from completing nuclear division and subsequent cytokinesis, which halts the proliferation of the cancer cells.
(b) [Max 4 marks] - Produces genetically identical daughter cells [1] - Growth: increases cell number / tissue size [1] - Repair of damaged tissues / replacement of dead or worn-out cells [1] (Reject: cell repair) - Asexual reproduction to produce identical offspring [1] - Clonal expansion in immune response / production of memory or plasma cells [1]
(c) [Max 3 marks] - Spindle fibres cannot shorten / depolymerise [1] - Sister chromatids cannot separate / be pulled to opposite poles [1] - Mitosis is arrested at metaphase / anaphase [1] - Cytokinesis / cell division cannot occur / prevents cell proliferation [1]
題目 6 · structured
10 分
(a) Distinguish between the mechanisms of a competitive inhibitor and a non-competitive inhibitor. [4]
(b) Explain why increasing substrate concentration can overcome the effect of a competitive inhibitor but has no effect on a non-competitive inhibitor. [4]
(c) Describe the *induced-fit hypothesis* of enzyme action. [2]
查看答案詳解收起答案詳解
解題
(a) A competitive inhibitor has a molecular structure similar to that of the substrate and binds directly to the active site, temporarily blocking the substrate from entering. In contrast, a non-competitive inhibitor does not resemble the substrate and binds to an allosteric site (a site other than the active site). This binding causes a conformational change in the enzyme's tertiary structure, altering the shape of the active site so that the substrate can no longer fit.
(b) In competitive inhibition, both substrate and inhibitor molecules compete for the same active site. If substrate concentration is high, the probability of a substrate molecule colliding with and binding to an active site is much greater than that of the inhibitor. This allows the maximum rate of reaction (\(V_{max}\)) to be reached. With non-competitive inhibition, the inhibitor binds to the allosteric site independently of substrate concentration. This permanently or temporarily deactivates the affected enzymes, reducing the total concentration of functional enzymes, meaning increasing substrate concentration cannot restore the maximum rate of reaction.
(c) The induced-fit hypothesis states that the active site of an enzyme is not fully complementary to the substrate initially. When the substrate collides with the active site, it induces a conformational change in the enzyme molecule, causing the active site to mold precisely around the substrate. This conformation strains chemical bonds within the substrate, lowering the activation energy needed for the reaction to proceed.
評分準則
(a) [Max 4 marks] - Competitive inhibitor has a similar structure/shape to the substrate [1] - Competitive inhibitor binds to the active site [1] - Non-competitive inhibitor binds to an allosteric site / site other than active site [1] - Non-competitive inhibitor alters the tertiary structure / changes shape of active site [1] - Non-competitive inhibitor does not compete with the substrate [1]
(b) [Max 4 marks] - Substrate and competitive inhibitor compete for the same active site [1] - At high substrate concentration, the substrate outcompetes the inhibitor [1] - More enzyme-substrate complexes form / \(V_{max}\) can be reached [1] - Non-competitive inhibitor alters the active site shape regardless of substrate presence [1] - Substrate cannot bind to the altered active site [1] - Maximum rate of reaction (\(V_{max}\)) is reduced / cannot be achieved [1]
(c) [Max 2 marks] - Active site is not initially fully complementary to the substrate [1] - Active site changes shape / molds around the substrate when it binds [1] - Puts strain on substrate bonds to lower activation energy [1]
Paper 33 (Advanced Practical Skills)
Answer all questions. You will need the materials and apparatus listed in the confidential instructions.
2 題目 · 40 分
題目 1 · Practical Investigations & Calculations
20 分
In this investigation, you will investigate the effect of substrate concentration on the rate of oxygen production by the enzyme catalase found in yeast. You are provided with: H, a 6.0% hydrogen peroxide solution; Y, a 5.0% active yeast suspension; and W, distilled water. (a) (i) You are required to prepare a two-fold serial dilution of the 6.0% hydrogen peroxide solution H to obtain four further concentrations: 3.0%, 1.5%, 0.75%, and 0.375%. Complete a table showing the volume of H and distilled water W needed to prepare 20 cm3 of each of these four concentrations. [4] (ii) You will carry out the investigation using yeast-impregnated paper discs. When a paper disc soaked in yeast suspension Y is placed in hydrogen peroxide, oxygen bubbles form on its surface, causing it to float. The time taken (t) for the disc to rise to the surface is recorded. Design and draw a table to record your results for all five concentrations of hydrogen peroxide (6.0%, 3.0%, 1.5%, 0.75%, 0.375%), including trials, mean time (t), and mean rate of reaction calculated as 1/t. [6] (iii) Identify one significant source of experimental error in this procedure and suggest how the apparatus or method could be modified to reduce this error. [3] (iv) A student performed three trials at the 3.0% substrate concentration and obtained the following times: 12.4 seconds, 11.8 seconds, and 12.1 seconds. Calculate the mean rate of reaction for this concentration. Express your answer in standard form to three significant figures, with appropriate units. Show your working. [3] (v) Explain the expected shape of the curve if the rate of reaction were plotted against substrate concentration from 0.375% to 6.0%. [4]
查看答案詳解收起答案詳解
解題
(a) (i) Serial dilution table: To make 20 cm3 of each concentration: For 3.0%: 10 cm3 of 6.0% H + 10 cm3 of W. For 1.5%: 10 cm3 of 3.0% H + 10 cm3 of W. For 0.75%: 10 cm3 of 1.5% H + 10 cm3 of W. For 0.375%: 10 cm3 of 0.75% H + 10 cm3 of W. (ii) Results table: The table must include: Headings for Substrate Concentration (%), Time taken for disc to rise (t) in seconds for Trial 1, Trial 2, Mean Time (t) in seconds, and Mean Rate (1/t) in s-1. Example values: 6.0%: mean time 8.5 s, rate 0.118 s-1; 3.0%: mean time 12.1 s, rate 0.0826 s-1; 1.5%: mean time 24.5 s, rate 0.0408 s-1; 0.75%: mean time 48.0 s, rate 0.0208 s-1; 0.375%: mean time 95.0 s, rate 0.0105 s-1. (iii) Source of error: Difficulty in standardizing the amount of yeast suspension on the paper disc (variation in blotting). Modification: Use a micropipette to apply a fixed volume (e.g., 20 microliters) of yeast suspension Y directly to each dry paper disc. (iv) Calculation: Mean time = (12.4 + 11.8 + 12.1) / 3 = 12.1 s. Mean rate = 1 / 12.1 = 0.08264 s-1. Expressed in standard form to 3 s.f. is 8.26 x 10^-2 s-1. (v) Explanation of curve: At lower substrate concentrations (0.375% to 1.5%), the rate of reaction increases proportionally with substrate concentration. Substrate concentration is the limiting factor. Active sites of catalase are not fully saturated. At higher substrate concentrations (3.0% to 6.0%), the rate of reaction levels off (reaches Vmax) because enzyme concentration becomes the limiting factor. All active sites are occupied (fully saturated) and enzyme-substrate complexes are forming at maximum rate.
評分準則
(a) (i) [4 marks] 1. Correct concentrations listed (3.0%, 1.5%, 0.75%, 0.375%). 2. Correct volumes of H (10 cm3 for each step). 3. Correct volumes of W (10 cm3 for each step). 4. Clear table format with appropriate headers and units. (ii) [6 marks] 1. Fully formatted table with cell lines drawn and headers clearly labelled with units (Concentration / %, Time / s, Rate / s-1). 2. Substrate concentrations in order (ascending or descending). 3. Raw times recorded to nearest whole second or 0.1 s. 4. Mean times calculated correctly from raw data. 5. Rates calculated correctly to 3 significant figures. 6. Consistent decimal places for all times. (iii) [3 marks] 1. Identifies error: variation in yeast volume/saturation of disc OR delay in starting stopwatch when disc hits liquid. 2. Explains effect of error: affects consistency of catalase concentration. 3. Suggests valid modification: use micropipette to add exact volume of yeast / use an automated release mechanism. (iv) [3 marks] 1. Show working: Mean time = 12.1 s. 2. Accuracy: 8.26 x 10^-2. 3. Units: s-1. (v) [4 marks] 1. Initial steep increase because substrate concentration is limiting / active sites available. 2. More frequent successful collisions / more enzyme-substrate complexes formed. 3. Plateaus at high concentration because enzyme concentration is limiting / active sites saturated. 4. Mention of Vmax or constant rate at maximum.
題目 2 · Practical Investigations & Calculations
20 分
In this investigation, you will investigate osmosis in plant tissue. You are provided with: S, a 1.0 mol dm-3 sucrose solution; P, fresh potato cylinders cut to approximately 50 mm in length; and W, distilled water. (a) (i) You are required to prepare a proportional dilution of the 1.0 mol dm-3 sucrose solution S to obtain four further concentrations: 0.8, 0.6, 0.4, and 0.2 mol dm-3. Complete a table showing the volume of S and distilled water W needed to prepare 20 cm3 of each of these four concentrations, alongside the original 1.0 mol dm-3 solution. [4] (ii) Describe a step-by-step method you would use to determine the concentration of sucrose that is isotonic to the potato tissue by measuring the change in length of the potato cylinders. [5] (iii) Present a results table showing hypothetical results for this experiment, including initial length (mm), final length (mm), change in length (mm), and percentage change in length (%) for all five sucrose concentrations. [5] (iv) Explain how you would use a graph of percentage change in length against sucrose concentration to determine the solute potential of the potato tissue. [2] (v) A student used a standard metric ruler with 1 mm divisions to measure the initial 50 mm length of a potato cylinder. Calculate the percentage uncertainty in this measurement, assuming two measurements are taken (one at each end of the cylinder). Show your working. [4]
查看答案詳解收起答案詳解
解題
(a) (i) Proportional dilution table: For 1.0 mol dm-3: 20 cm3 of S + 0 cm3 of W. For 0.8 mol dm-3: 16 cm3 of S + 4 cm3 of W. For 0.6 mol dm-3: 12 cm3 of S + 8 cm3 of W. For 0.4 mol dm-3: 8 cm3 of S + 12 cm3 of W. For 0.2 mol dm-3: 4 cm3 of S + 16 cm3 of W. (ii) Step-by-step method: 1. Cut five potato cylinders using a cork borer and trim each to exactly 50 mm using a scalpel and ruler. 2. Place 20 cm3 of each sucrose concentration into separate labelled boiling tubes. 3. Record the initial length of each potato cylinder. 4. Place one cylinder into each tube and leave for at least 30 minutes. 5. Remove the cylinders, gently blot dry with a paper towel to remove excess surface water, and measure the final length. (iii) Results table: Example values (initial length 50.0 mm): 0.2 mol dm-3: final 52.0 mm, change +2.0 mm, % change +4.0%; 0.4 mol dm-3: final 50.5 mm, change +0.5 mm, % change +1.0%; 0.6 mol dm-3: final 49.0 mm, change -1.0 mm, % change -2.0%; 0.8 mol dm-3: final 48.0 mm, change -2.0 mm, % change -4.0%; 1.0 mol dm-3: final 47.0 mm, change -3.0 mm, % change -6.0%. (iv) Plot percentage change in length against sucrose concentration. Draw a line of best fit. The point where the line intersects the x-axis (0% change in length) represents the concentration of sucrose that is isotonic. Use this concentration to look up the equivalent solute potential from standard tables. (v) Calculation: A ruler has an uncertainty of +/- 0.5 mm at each end. Total uncertainty for a reading of length (difference between two marks) is 2 * 0.5 = +/- 1.0 mm. Percentage uncertainty = (1.0 mm / 50.0 mm) * 100% = 2.0%.
評分準則
(a) (i) [4 marks] 1. Table has clear headers: Sucrose concentration / mol dm-3, Volume of S / cm3, Volume of W / cm3. 2. All volumes calculated correctly to make 20 cm3 total. 3. Volumes are consistent with proportional dilution. 4. Table includes all 5 concentrations (including 1.0 mol dm-3). (ii) [5 marks] 1. Describes cutting cylinders to a standard initial length (e.g., 50 mm) using cork borer and scalpel. 2. Mentions immersion in specified volumes (e.g., 20 cm3) of sucrose solutions. 3. Specifies leaving for a minimum of 30 minutes. 4. Describes blotting cylinders dry with paper towel before re-measuring. 5. Mentions measuring final length and calculating change. (iii) [5 marks] 1. Heading row has units (Concentration / mol dm-3, Initial length / mm, Final length / mm, Change / mm, Percentage change / %). 2. All raw measurements given to the same precision (e.g., to 0.5 mm or 1 mm). 3. Shows both positive changes (swelling at low concentration) and negative changes (shrinkage at high concentration). 4. Correct calculation of percentage changes. 5. Column for signs (+/-) for change and percentage change. (iv) [2 marks] 1. Identify x-intercept (where y = 0 or no change in length) as the isotonic concentration. 2. Reference to converting isotonic concentration to solute potential using calibration curve or table. (v) [4 marks] 1. Identifies uncertainty of a single reading as +/- 0.5 mm. 2. Recognizes that measuring a length involves two readings (uncertainty of +/- 1.0 mm). 3. Shows working: (1.0 / 50.0) * 100. 4. Correct final answer of 2.0% (accept 2%).
Paper 43 (A Level Structured)
Answer all ten structured questions in the spaces provided. Calculators are permitted.
10 題目 · 100 分
題目 1 · A Level Structured
10 分
An investigation was conducted into the activity of the enzyme ribulose bisphosphate carboxylase-oxygenase (rubisco) in the leaves of the halophyte plant, *Salicornia europaea*, under different salinity treatments.
(a) Describe the role of ribulose bisphosphate (RuBP) in the light-independent stage of photosynthesis. [3]
(b) Rubisco can act as both a carboxylase and an oxygenase. Explain why an increase in oxygen concentration relative to carbon dioxide concentration decreases the rate of carbon fixation in C3 plants. [4]
(c) High salinity conditions often trigger stomatal closure in plants. Suggest and explain the effect of prolonged stomatal closure on the concentrations of RuBP and glycerate 3-phosphate (GP) in the stroma of the chloroplast. [3]
查看答案詳解收起答案詳解
解題
(a) RuBP (a 5-carbon sugar phosphate) acts as the primary carbon dioxide acceptor in the stroma during the light-independent stage of photosynthesis. It combines with carbon dioxide in a reaction catalysed by the enzyme rubisco (carbon fixation) to form an unstable 6-carbon intermediate, which immediately splits into two molecules of the 3-carbon compound glycerate 3-phosphate (GP).
(b) Oxygen acts as a competitive inhibitor to carbon dioxide for the active site of the enzyme rubisco. When the concentration of oxygen increases relative to carbon dioxide, oxygen binds to the active site instead of carbon dioxide (the oxygenase activity of rubisco). This leads to a process called photorespiration. Consequently, less RuBP is carboxylated (combined with carbon dioxide), which directly reduces the rate of GP production and overall carbon fixation.
(c) Stomatal closure reduces the diffusion of carbon dioxide into the leaf, causing a decrease in the concentration of carbon dioxide in the stroma. As a result, there is less carbon dioxide available to combine with RuBP, causing RuBP to accumulate so its concentration increases. Conversely, GP continues to be converted into triose phosphate (TP) using ATP and reduced NADP from the light-dependent stage, but is not replenished by carbon fixation, leading to a decrease in the concentration of GP.
評分準則
(a) Max 3 marks: 1. RuBP is a 5C compound / CO2 acceptor; 2. Combines with CO2 / undergoes carbon fixation; 3. Catalysed by the enzyme rubisco; 4. Produces an unstable 6C intermediate which splits into two molecules of GP (3C);
(b) Max 4 marks: 1. Oxygen is a competitive inhibitor / has a complementary shape to the active site of rubisco; 2. Oxygen binds to the active site of rubisco / oxygenase activity occurs; 3. Less RuBP combines with CO2; 4. Leads to photorespiration / waste of RuBP; 5. Rate of GP production / carbon fixation decreases;
(c) Max 3 marks: 1. Stomatal closure decreases carbon dioxide entry / concentration in stroma; 2. RuBP concentration increases / accumulates because less CO2 is available to react with it; 3. GP concentration decreases because GP is still converted to TP (using ATP and reduced NADP) but is not being produced by carbon fixation;
題目 2 · A Level Structured
10 分
Osmoregulation is an important homeostatic mechanism in mammals that maintains the water potential of the blood within narrow limits.
(a) State the precise location of osmoreceptors in the mammalian brain and outline how they detect a decrease in the water potential of the blood. [3]
(b) Describe the cell signaling events that occur in a collecting duct epithelial cell after antidiuretic hormone (ADH) binds to its cell surface receptor, leading to an increase in water reabsorption. [5]
(c) Nephrogenic diabetes insipidus is a condition where the kidneys fail to respond to ADH. Suggest how a mutation in the gene encoding the ADH receptor can lead to the production of large volumes of dilute urine. [2]
查看答案詳解收起答案詳解
解題
(a) Osmoreceptors are located in the hypothalamus of the brain. When the water potential of the blood decreases, water moves out of the osmoreceptor cells into the blood by osmosis, down a water potential gradient. This causes the osmoreceptor cells to shrink/lose volume, which stimulates them to generate nerve impulses that travel to the posterior pituitary gland.
(b) The binding of ADH to its specific cell surface receptor on the basolateral membrane of collecting duct cells activates a G-protein. This active G-protein stimulates the enzyme adenylyl cyclase, which catalyses the conversion of ATP to cyclic AMP (cAMP). cAMP acts as a second messenger, activating a protein kinase intracellular cascade. This cascade stimulates vesicles containing aquaporin-2 (AQP2) proteins to move towards and fuse with the apical (luminal) membrane of the cell via exocytosis. This inserts more aquaporins into the apical membrane, increasing its permeability to water.
(c) A mutation in the gene encoding the ADH receptor changes the primary structure of the receptor protein, which alters its tertiary structure and the shape of its binding site. Consequently, ADH can no longer bind to the receptor, preventing the activation of the G-protein and intracellular signaling pathway. Aquaporins are not inserted into the apical membranes of collecting duct cells, leaving them impermeable to water, so water cannot be reabsorbed by osmosis and is excreted as a large volume of dilute urine.
評分準則
(a) Max 3 marks: 1. Located in the hypothalamus; 2. Detect decrease in blood water potential; 3. Water moves out of osmoreceptor cells by osmosis (down water potential gradient); 4. Osmoreceptors shrink / lose volume, triggering nerve impulses (to posterior pituitary);
(b) Max 5 marks: 1. ADH binds to receptor on cell surface membrane (basolateral membrane); 2. Activates G-protein; 3. G-protein activates adenylyl cyclase; 4. Adenylyl cyclase converts ATP to cyclic AMP / cAMP (second messenger); 5. cAMP activates protein kinase (A) cascade; 6. Vesicles containing aquaporins / AQP2 move to and fuse with apical / luminal membrane (exocytosis); 7. Increases permeability of membrane to water;
(c) Max 2 marks: 1. Altered tertiary structure of ADH receptor / change in shape of binding site so ADH cannot bind; 2. No intracellular cascade / no cAMP production / no aquaporins inserted into membrane; 3. Collecting duct remains impermeable to water, so water is not reabsorbed and is excreted;
題目 3 · A Level Structured
10 分
Auxins, such as indol-3-acetic acid (IAA), play a vital role in controlling plant growth and coordination.
(a) Outline how auxin is redistributed in a shoot tip during a phototropic response to unilateral light. [3]
(b) Explain the "acid growth hypothesis" of auxin-induced cell elongation, describing the roles of proton pumps, expansins, and cell wall structure. [5]
(c) Suggest why treating plant tissue with an inhibitor of translation (protein synthesis) prevents long-term auxin-induced cell elongation, even if proton pump activity is maintained initially. [2]
查看答案詳解收起答案詳解
解題
(a) Auxin (IAA) is synthesized in the apical meristem of the shoot tip. When unilateral light shines on one side of the shoot, auxin is transported laterally away from the illuminated side towards the shaded side of the shoot tip. This active lateral transport results in a higher concentration of auxin accumulating on the shaded side of the shoot compared to the illuminated side.
(b) Auxin binds to specific receptors (such as ABP1) on the cell surface membrane of target cells. This stimulates proton pumps (H+-ATPases) in the cell surface membrane to actively transport H+ ions from the cytoplasm into the cell wall (apoplast). The accumulation of H+ ions lowers the pH of the cell wall. This acidic environment activates proteins called expansins, which break the non-covalent (hydrogen) bonds between cellulose microfibrils and other cell wall matrix components (like hemicelluloses). The cell wall becomes loose and extensible. Turgor pressure inside the vacuole/cell then stretches the loosened cell wall, leading to cell elongation.
(c) Long-term cell elongation requires the continuous production of new cell wall materials (such as cellulose, hemicellulose, and structural proteins) to maintain cell wall thickness and mechanical integrity as the cell expands. Inhibiting translation prevents the synthesis of these structural proteins and the enzymes required for cell wall synthesis, as well as preventing the replacement of proton pump proteins, which stops sustained elongation.
評分準則
(a) Max 3 marks: 1. Auxin / IAA synthesized in the shoot tip / apical meristem; 2. Unilateral light causes active lateral transport of auxin; 3. Auxin moves from the illuminated side to the shaded side; 4. Creates a higher concentration of auxin on the shaded side;
(b) Max 5 marks: 1. Auxin binds to cell surface receptors; 2. Stimulates proton pumps / H+-ATPases in cell membrane; 3. H+ ions actively pumped into the cell wall / apoplast, lowering the pH; 4. Low pH activates expansins; 5. Expansins break hydrogen bonds between cellulose microfibrils and hemicelluloses / cell wall matrix; 6. Cell wall becomes loose / flexible / extensible; 7. Turgor pressure drives water entry and stretches / expands the cell;
(c) Max 2 marks: 1. Long-term growth requires synthesis of new cell wall components / proteins / enzymes; 2. Inhibitor prevents synthesis of structural proteins / enzymes needed to make cell wall thicker / stronger; 3. Prevents replacement / synthesis of proton pump proteins;
題目 4 · A Level Structured
10 分
Polymerase chain reaction (PCR) is an essential technique in genetic technology used to amplify specific target DNA sequences.
(a) State the precise role of each of the following components in a PCR reaction mixture: (i) Taq polymerase [1] (ii) DNA primers [2]
(b) Describe the temperature changes during one complete cycle of PCR and explain the purpose of each temperature step. [4]
(c) Microarray analysis is often used to compare gene expression profiles between different tissue samples. Outline how a microarray is used to detect differences in the expression of specific genes between two tissue samples. [3]
查看答案詳解收起答案詳解
解題
(a) (i) Taq polymerase is a thermostable DNA polymerase that synthesizes complementary DNA strands by adding free deoxynucleoside triphosphates (dNTPs) to the 3' end of the primers. Because it is heat-tolerant, it does not denature at the high temperatures used in PCR. (ii) DNA primers are short, single-stranded sequences of DNA that are complementary to the sequences at the ends of the target DNA region. They bind (anneal) to the single-stranded DNA templates and provide the double-stranded starting point required for Taq polymerase to initiate DNA synthesis.
(b) One PCR cycle consists of three temperature steps: 1. Denaturation: The mixture is heated to 92-95 °C to break the hydrogen bonds between the complementary base pairs of the double-stranded template DNA, separating it into single strands. 2. Annealing: The temperature is lowered to 50-65 °C to allow the DNA primers to form hydrogen bonds with their complementary sequences on the single-stranded DNA templates. 3. Extension: The temperature is raised to 70-75 °C, which is the optimum temperature for Taq polymerase to rapidly synthesize the complementary DNA strands by adding nucleotides.
(c) mRNA is extracted from both tissue samples and reverse transcribed to produce complementary DNA (cDNA). The cDNA from each tissue is labeled with a different colored fluorescent dye (e.g., green for tissue A, red for tissue B). The labeled cDNA samples are mixed and applied to the microarray slide, where they hybridize with matching single-stranded DNA probes fixed to the wells. The microarray is washed to remove unbound cDNA and scanned with a laser. The intensity and color of fluorescence in each spot reveal which genes are expressed and their relative abundance in each tissue.
評分準則
(a) (i) Max 1 mark: 1. Thermostable enzyme that synthesizes complementary DNA strands / adds free nucleotides to the 3' end of primers (without denaturing at high temperatures); (ii) Max 2 marks: 1. Short, single-stranded DNA sequences complementary to the target DNA; 2. Bind / anneal to template strands to provide a double-stranded starting region for Taq polymerase to bind / start synthesis;
(b) Max 4 marks: 1. 92-95 °C: Denaturation to break hydrogen bonds between base pairs and separate DNA into single strands; 2. 50-65 °C: Annealing to allow primers to bind / form hydrogen bonds with complementary template sequences; 3. 70-75 °C: Extension, which is the optimum temperature for Taq polymerase activity to synthesize new DNA strands;
(c) Max 3 marks: 1. mRNA extracted and reverse transcribed to cDNA; 2. cDNA samples labeled with different fluorescent dyes (e.g., red and green); 3. Labeled cDNA hybridizes with single-stranded probes on the microarray; 4. Wash to remove unhybridized cDNA and scan with laser to detect relative intensity of fluorescence (indicating gene expression levels);
題目 5 · A Level Structured
10 分
The grass species *Agrostis capillaris* is able to colonize areas contaminated with heavy metals, such as old copper mine tailings, where other plants fail to survive.
(a) Explain how natural selection has led to a high frequency of heavy metal-tolerant alleles in the population of *A. capillaris* growing on the mine tailings compared to the adjacent uncontaminated pasture. [5]
(b) Even though the populations of *A. capillaris* on the mine tailings and the adjacent pasture are geographically contiguous (touching), gene flow between them is limited, leading to reproductive isolation. Explain how reproductive isolation could arise between these two populations. [3]
(c) State two conditions that must be met for a population to remain in Hardy-Weinberg equilibrium. [2]
查看答案詳解收起答案詳解
解題
(a) Within the original *A. capillaris* population, genetic variation existed due to random gene mutations, resulting in some individuals possessing alleles that confer tolerance to high concentrations of heavy metals. On the mine tailings, the presence of toxic heavy metals acts as a strong selective pressure. Non-tolerant plants are poisoned and die, while individuals possessing the metal-tolerant alleles survive and grow successfully. These surviving plants reproduce sexually and pass on the advantageous metal-tolerant alleles to their offspring. Over many generations, this directional selection increases the allele frequency for heavy metal tolerance on the mine tailings, whereas on the uncontaminated pasture, there is no such selective pressure (and tolerance alleles may even carry a metabolic cost, keeping their frequency low).
(b) Reproductive isolation can arise through directional selection favoring different flowering times (allochronic isolation) in the two environments; for example, metal-tolerant plants may flower earlier or later than non-tolerant plants, preventing cross-pollination. Additionally, selection against hybrids occurs: hybrid offspring (possessing intermediate tolerance) are less fit on the mine tailings than fully tolerant plants, and less fit on the pasture than non-tolerant plants. This selective disadvantage reduces gene flow, leading to physiological, behavioral, or genetic barriers that isolate the gene pools.
(c) Two conditions for Hardy-Weinberg equilibrium are: 1. No mutation (no new alleles are created). 2. No natural selection (all genotypes have equal survival and reproductive success). 3. The population is extremely large. 4. Mating is completely random. 5. No migration / gene flow in or out of the population. (Any two)
評分準則
(a) Max 5 marks: 1. Variation exists in original population due to random mutation; 2. Some individuals possess alleles for heavy metal tolerance; 3. Toxic heavy metals in mine tailings act as a selection pressure; 4. Non-tolerant plants are selected against / die, while tolerant plants survive / show selective advantage; 5. Tolerant plants reproduce and pass on the advantageous allele to offspring; 6. Frequency of metal-tolerant allele increases over generations (directional selection); 7. On uncontaminated pasture, there is no selection pressure for tolerance (or tolerance alleles have a selective disadvantage/metabolic cost);
(b) Max 3 marks: 1. Divergence in flowering times (ecological/temporal/allochronic isolation) prevents cross-pollination; 2. Selection against hybrids / hybrid offspring are less adapted to either environment (reduced fitness); 3. Disruption of gene flow leads to genetic divergence between the two populations;
(c) Max 2 marks (any two from): 1. No mutation; 2. No selection pressure / all genotypes have equal survival; 3. Large population size; 4. Random mating; 5. No migration / gene flow;
題目 6 · A Level Structured
10 分
An ecological study was carried out to compare the biodiversity of ground-dwelling beetles in a managed pine plantation and an ancient semi-natural deciduous woodland using pitfall traps.
(a) Distinguish between species richness and species evenness. [2]
(b) Simpson’s Index of Diversity (\(D\)) is calculated using the formula:
\[D = 1 - \sum \left(\frac{n}{N}\right)^2\]
where \(n\) is the number of individuals of a particular species and \(N\) is the total number of individuals of all species.
In the managed pine plantation, five species of beetles were recorded with the following individual counts: - Species A: 120 - Species B: 15 - Species C: 8 - Species D: 4 - Species E: 3
(i) Calculate the value of Simpson's Index of Diversity (\(D\)) for this plantation. Show your working. [3] (ii) State what a value of \(D\) close to 0 indicates about the stability and diversity of this ecosystem. [1]
(c) Describe how a belt transect and quadrats could be used alongside a systematic sampling method to investigate changes in the abundance and distribution of plant species across the transition zone (ecotone) between the pine plantation and the deciduous woodland. [4]
查看答案詳解收起答案詳解
解題
(a) Species richness is the total number of different species present in a community or habitat. Species evenness is a measure of the relative abundance or proportional numbers of individuals of each of these species in the habitat.
(b) (i) First, calculate the total number of individuals (\(N\)): \(N = 120 + 15 + 8 + 4 + 3 = 150\).
Next, calculate \(\left(\frac{n}{N}\right)^2\) for each species: - Species A: \(\left(\frac{120}{150}\right)^2 = (0.8)^2 = 0.6400\) - Species B: \(\left(\frac{15}{150}\right)^2 = (0.1)^2 = 0.0100\) - Species C: \(\left(\frac{8}{150}\right)^2 = (0.0533)^2 = 0.00284\) - Species D: \(\left(\frac{4}{150}\right)^2 = (0.0267)^2 = 0.00071\) - Species E: \(\left(\frac{3}{150}\right)^2 = (0.02)^2 = 0.00040\)
Sum of \(\left(\frac{n}{N}\right)^2\): \(\sum \left(\frac{n}{N}\right)^2 = 0.6400 + 0.0100 + 0.00284 + 0.00071 + 0.00040 = 0.65395\)
Subtract this sum from 1: \(D = 1 - 0.65395 = 0.346\) (accept 0.35)
(ii) A value of \(D\) close to 0 indicates low species diversity, dominated by a single or very few species, meaning the ecosystem is unstable and highly vulnerable to environmental changes.
(c) Lay a long tape measure from the pine plantation, through the transition zone, and into the deciduous woodland to form a line/belt transect. Place a quadrat of standard size (e.g., 0.25 m²) at regular, systematic intervals along the tape measure (e.g., every 5 meters). Inside each quadrat, identify the plant species present and estimate their abundance using percentage cover, species frequency, or direct counts of individual plants. To ensure reliability and representative data, repeat the transect line multiple times parallel to the first one and calculate mean values of abundance for each interval distance.
評分準則
(a) Max 2 marks: 1. Species richness: total number of different species in a community / habitat; 2. Species evenness: relative abundance / proportion of individuals of each species;
(b) (i) Max 3 marks: 1. Correct calculation of N = 150; 2. Correct calculation of sum of squares: \(\sum (n/N)^2 = 0.654\) (accept working values); 3. Final answer: \(D = 0.346\) or \(0.35\) (allow ecf from step 2); (ii) Max 1 mark: 1. Indicates low diversity / dominated by one species AND unstable / vulnerable to change;
(c) Max 4 marks: 1. Lay tape measure from plantation into woodland to form a transect line; 2. Place quadrats at regular / systematic intervals (e.g. every 5m) along the tape measure; 3. Identify plant species and record abundance (e.g. using percentage cover / direct count / local frequency) in each quadrat; 4. Replicate / lay multiple parallel transects to obtain mean / representative values;
題目 7 · A Level Structured
10 分
Respiration consists of a series of metabolic pathways that generate ATP to drive cellular processes.
(a) Outline the pathway of glycolysis, indicating precisely where ATP is used and where ATP is produced. [4]
(b) Explain how the link reaction and the Krebs cycle provide substrates for the reactions of the electron transport chain (ETC). [3]
(c) Rotenone is a chemical that inhibits Complex I of the electron transport chain, stopping the oxidation of reduced NAD (NADH). Explain why rotenone leads to a significant decrease in ATP synthesis by oxidative phosphorylation, and suggest why some ATP can still be synthesized. [3]
查看答案詳解收起答案詳解
解題
(a) Glycolysis occurs in the cytoplasm. First, glucose (6C) is phosphorylated by ATP to form glucose phosphate, which is then converted and phosphorylated again using another ATP to form fructose 1,6-bisphosphate (using 2 ATP in total). This unstable 6C compound splits into two molecules of triose phosphate (3C). Each triose phosphate is oxidized by losing hydrogen to NAD (forming reduced NAD). Substrate-level phosphorylation then occurs, where phosphate groups are transferred to ADP, producing 4 ATP molecules in total and converting triose phosphate into two molecules of pyruvate (3C). This results in a net yield of 2 ATP and 2 reduced NAD.
(b) The link reaction and the Krebs cycle oxidize organic molecules, removing hydrogen atoms (protons and high-energy electrons). These hydrogens are accepted by the coenzymes NAD and FAD to form reduced NAD (NADH) and reduced FAD (FADH2). These reduced coenzymes then act as the substrates for the electron transport chain; they travel to the inner mitochondrial membrane where they release their electrons and protons to the electron carriers.
(c) Complex I is the specific membrane protein that accepts electrons from reduced NAD. When Complex I is inhibited by rotenone, reduced NAD cannot be oxidized, so its electrons cannot enter the electron transport chain. This stops proton pumping at Complex I, severely reducing the electrochemical proton gradient across the inner mitochondrial membrane, so less proton flow passes through ATP synthase to produce ATP. However, some ATP can still be synthesized because reduced FAD (FADH2) bypasses Complex I, delivering its electrons directly to Complex II. This allows electron flow and proton pumping to continue at subsequent complexes (Complex III and IV), generating a smaller proton gradient that can still drive limited ATP synthesis.
評分準則
(a) Max 4 marks: 1. Glucose phosphorylated by ATP to form glucose phosphate, then fructose 1,6-bisphosphate (2 ATP used); 2. Fructose 1,6-bisphosphate splits into two triose phosphate (3C) molecules; 3. Triose phosphate is oxidized / dehydrogenated to form pyruvate (3C); 4. Hydrogen is accepted by NAD to form reduced NAD / NADH; 5. Phosphates transferred to ADP to form ATP via substrate-level phosphorylation (4 ATP produced / net gain of 2 ATP);
(b) Max 3 marks: 1. Link reaction and Krebs cycle dehydrogenate / oxidize substrates; 2. NAD and FAD are reduced / accept hydrogen atoms (electrons and protons); 3. Reduced NAD and reduced FAD diffuse to the inner mitochondrial membrane / cristae to deliver electrons and protons to the ETC;
(c) Max 3 marks: 1. Inhibition of Complex I prevents oxidation of reduced NAD, stopping its electron flow; 2. Less protons are pumped across the inner membrane / proton gradient is reduced / proton motive force decreases; 3. Reduced FAD (FADH2) donates its electrons to Complex II / bypasses Complex I; 4. Electrons from FADH2 still flow through the remaining ETC, allowing some proton pumping and ATP synthesis via ATP synthase;
題目 8 · A Level Structured
10 分
The regulation of gene expression in prokaryotes can be illustrated by the *lac* operon in *Escherichia coli*.
(a) State the precise functions of the structural genes *lacZ* and *lacY* in the *lac* operon. [2]
(b) Describe the state of the *lac* operon in *E. coli* when: (i) Lactose is absent. [3] (ii) Lactose is present and glucose is absent. [3]
(c) Distinguish between the roles of structural genes and regulatory genes. [2]
查看答案詳解收起答案詳解
解題
(a) *lacZ* encodes the enzyme beta-galactosidase, which hydrolyses the disaccharide lactose into the monosaccharides glucose and galactose. *lacY* encodes the membrane-bound transport protein lactose permease, which facilitates the uptake of lactose from the surrounding medium into the bacterial cell.
(b) (i) When lactose is absent, the regulatory gene *lacI* is continuously transcribed and translated to produce an active repressor protein. This repressor protein binds directly to the operator region of the operon. Because the repressor is bound to the operator, it physically blocks RNA polymerase from binding to the promoter and moving along the DNA. Consequently, transcription of the structural genes (*lacZ*, *lacY*, and *lacA*) is switched off (prevented). (ii) When lactose is present, it is converted to allolactose, which binds to the repressor protein. This binding causes a conformational (shape) change in the repressor protein, rendering it inactive so that it can no longer bind to the operator. RNA polymerase can now bind to the promoter and transcribe the structural genes. Since glucose is absent, cyclic AMP (cAMP) levels in the cell are high, which allows cAMP to bind to the catabolite activator protein (CAP). This active complex binds to the promoter, facilitating the binding of RNA polymerase and stimulating transcription of the operon at a maximum rate.
(c) Structural genes code for proteins or enzymes that play a direct structural, catalytic, or transport role within the cell (and do not directly regulate the expression of other genes). Regulatory genes code for proteins, such as transcription factors, repressors, or activators, that control the transcription of other genes by binding to specific DNA sequences (like operators or promoters) or to other regulatory molecules.
評分準則
(a) Max 2 marks: 1. *lacZ* codes for beta-galactosidase, which hydrolyses lactose to glucose and galactose; 2. *lacY* codes for lactose permease, which transports lactose into the cell;
(b) (i) Max 3 marks: 1. Regulatory gene (*lacI*) is expressed / repressor protein is produced; 2. Repressor protein binds to the operator; 3. Prevents RNA polymerase from binding to promoter / transcribing structural genes; (ii) Max 3 marks: 1. Lactose / allolactose binds to the repressor protein; 2. Repressor protein changes shape / becomes inactive and detaches from operator; 3. RNA polymerase binds to promoter and transcribes structural genes; 4. Low glucose leads to high cAMP, which binds to CAP to increase/maximize transcription rate;
(c) Max 2 marks: 1. Structural genes: code for proteins with structural, transport, or metabolic/catalytic functions in the cell; 2. Regulatory genes: code for proteins (e.g. repressor/transcription factor) that control the expression of other genes;
題目 9 · A Level 結構題
10 分
Some plants and micro-organisms produce natural toxins that can disrupt aerobic respiration.
(a) Explain why nicotinamide adenine dinucleotide (NAD) and flavin adenine dinucleotide (FAD) are described as coenzymes in respiration. [3]
(b) Antimycin A is a piscicide (fish toxin) that acts as an inhibitor of Complex III in the mitochondrial electron transport chain.
(i) Describe the immediate effect of Antimycin A on the proton gradient across the inner mitochondrial membrane. [2]
(ii) Explain why the consumption of oxygen by a suspension of isolated mitochondria decreases dramatically when Antimycin A is added. [3]
(c) Yeast cells can switch between aerobic and anaerobic respiration depending on oxygen availability. Contrast the yield of ATP per mole of glucose under aerobic conditions with the yield of ATP under anaerobic conditions in yeast cells. [2]
查看答案詳解收起答案詳解
解題
(a) NAD and FAD act as coenzymes because they are non-protein organic molecules required for the functioning of respiratory enzymes (dehydrogenases). They carry hydrogen atoms (protons and electrons) removed from respiratory substrates during glycolysis, the link reaction, and the Krebs cycle, transferring them to the electron transport chain on the inner mitochondrial membrane where they are oxidised and recycled.
(b) (i) Antimycin A blocks the transfer of electrons through Complex III, stopping the pumping of protons from the mitochondrial matrix into the intermembrane space. As protons continue to flow back into the matrix through ATP synthase, the proton gradient across the inner mitochondrial membrane is rapidly depleted.
(ii) Oxygen acts as the terminal electron acceptor in the electron transport chain, where it combines with electrons and protons to form water. Since Antimycin A inhibits Complex III, the flow of electrons along the chain ceases. Consequently, Complex IV cannot transfer electrons to oxygen, preventing its reduction to water and causing oxygen consumption to drop dramatically.
(c) Under aerobic conditions, the complete oxidation of glucose via glycolysis, the link reaction, the Krebs cycle, and oxidative phosphorylation yields approximately 30 to 32 molecules of ATP per molecule of glucose. Under anaerobic conditions in yeast (ethanol fermentation), glycolysis is the only source of ATP, yielding only 2 molecules of ATP per molecule of glucose because the link reaction, Krebs cycle, and oxidative phosphorylation cannot occur.
評分準則
**Part (a) [Max 3]** 1. coenzymes are non-protein organic molecules necessary for enzyme activity; 2. accept / carry / transfer hydrogen atoms / protons and electrons / reducing equivalents; 3. from glycolysis / link reaction / Krebs cycle to the electron transport chain / cristae / inner membrane; 4. they are constantly recycled / regenerated (by being oxidised and reduced);
**Part (b)(i) [Max 2]** 1. stops / decreases the pumping of protons (\(H^+\)) from the matrix to the intermembrane space; 2. the proton gradient is reduced / depleted (as protons diffuse back into the matrix through ATP synthase);
**Part (b)(ii) [Max 3]** 1. oxygen is the final / terminal electron acceptor; 2. oxygen combines with electrons (from the ETC) and protons (\(H^+\)) to form water / \(H_2O\); 3. (with Complex III blocked) electron flow / transport along the ETC stops; 4. Complex IV cannot pass electrons to oxygen / oxygen is not reduced;
**Part (c) [Max 2]** 1. aerobic yield is much higher, around 30 to 32 ATP per glucose molecule (accept range 28 to 38); 2. anaerobic yield is much lower, only 2 ATP per glucose molecule; 3. anaerobic only involves glycolysis / substrate-level phosphorylation, whereas aerobic also involves oxidative phosphorylation / link / Krebs;
題目 10 · A Level 結構題
10 分
Maize (Zea mays) is an example of a C4 crop plant that exhibits high productivity in hot, dry climates.
(a) State the advantage of C4 photosynthesis over C3 photosynthesis in hot, dry environments. [2]
(b) Describe how the leaf anatomy of C4 plants is adapted to minimise photorespiration. [4]
(c) State the role of phosphoenolpyruvate (PEP) carboxylase in C4 plants and explain why its high affinity for carbon dioxide is advantageous. [4]
查看答案詳解收起答案詳解
解題
(a) The primary advantage of C4 photosynthesis is that high rates of photosynthesis can be maintained at high temperatures, high light intensities, and under dry conditions, whilst minimizing water loss and drastically reducing the wasteful process of photorespiration.
(b) C4 leaves display Kranz anatomy, characterized by concentric rings of mesophyll cells surrounding an inner ring of bundle sheath cells around the vascular bundle. The bundle sheath cells have thick walls that are impermeable to gases and contain specialized, often agranal chloroplasts. This spatial separation keeps oxygen produced by the light-dependent reactions in the mesophyll cells away from RuBisCO, which is localized strictly in the bundle sheath cells, thereby minimizing its oxygenase activity (photorespiration).
(c) PEP carboxylase catalyses the initial fixation of carbon dioxide (as bicarbonate ions) to phosphoenolpyruvate (PEP) in mesophyll cells, forming the 4-carbon compound oxaloacetate. Since PEP carboxylase has a very high affinity for carbon dioxide and does not bind oxygen, it can efficiently fix carbon dioxide even at very low concentrations (e.g., when stomata are partially closed to conserve water). This ensures a high, concentrated supply of carbon dioxide is continuously delivered to the bundle sheath cells, keeping RuBisCO saturated and preventing photorespiration.
評分準則
**Part (a) [Max 2]** 1. maintains high rates of photosynthesis at high temperature / high light intensity / dry conditions; 2. reduces / prevents photorespiration; 3. reduces water loss (as stomata can be partially closed without stopping carbon fixation);
**Part (b) [Max 4]** 1. Kranz anatomy / description of concentric rings of mesophyll and bundle sheath cells; 2. spatial separation of carbon dioxide fixation (in mesophyll) and Calvin cycle (in bundle sheath); 3. bundle sheath cells have thick walls / impermeable to gases (keeps oxygen out / carbon dioxide in); 4. chloroplasts in bundle sheath cells lack grana / are agranal, so no oxygen is generated by the light-dependent stage near RuBisCO; 5. RuBisCO is isolated from high oxygen concentrations;
**Part (c) [Max 4]** 1. PEP carboxylase catalyses reaction of carbon dioxide / bicarbonate with PEP to form oxaloacetate / malate / 4C compound (in mesophyll cells); 2. high affinity allows carbon dioxide to be fixed even at very low internal concentrations / when stomata are closed; 3. PEP carboxylase has no affinity for oxygen / does not bind oxygen (unlike RuBisCO); 4. ensures a high partial pressure / concentration of \(CO_2\) is maintained in bundle sheath cells (for RuBisCO);
Paper 53 (Planning, Analysis and Evaluation)
Answer all questions. Show your working in calculation steps and state null hypotheses where appropriate.
2 題目 · 30 分
題目 1 · Data Analysis & Methodology Design
15 分
A student investigated the effect of a competitive inhibitor, galactose, on the activity of the enzyme lactase, which hydrolyses lactose into glucose and galactose. The student was provided with: 10.0% stock solution of lactose, 2.0% stock solution of lactase, 5.0% stock solution of galactose (inhibitor), and distilled water. (a) Describe how the student could prepare 10 cm\(^3\) of each of the five concentrations of lactose solution: 2.0%, 4.0%, 6.0%, 8.0%, and 10.0%, using a simple dilution series from the 10.0% stock solution. [3] (b) Design a laboratory method to investigate the effect of galactose concentration on the rate of lactose hydrolysis. Your method should be detailed enough to allow another person to carry it out, and it must allow the student to determine if the inhibition is competitive by varying the substrate concentration. In your description, you must include: how the independent, dependent, and key controlled variables are managed, a reliable method for measuring the initial rate of reaction, and safety precautions. [6] (c) The student carried out the investigation and obtained the following results for the initial rate of reaction (\(v\)) at different lactose concentrations, in the presence and absence of 1.0% galactose inhibitor: Lactose concentration / % | Initial rate without inhibitor / a.u. | Initial rate with 1.0% galactose / a.u.: [1.0 | 12.5 | 5.2], [2.0 | 22.0 | 9.8], [4.0 | 35.5 | 18.1], [6.0 | 44.0 | 25.4], [8.0 | 48.2 | 32.1], [10.0 | 49.5 | 37.5], [15.0 | 50.0 | 45.0], [20.0 | 50.0 | 49.0]. (i) State the value of \(V_{\text{max}}\) for the reaction without inhibitor, and estimate the Michaelis-Menten constant (\(K_m\)) for the enzyme without inhibitor. Explain how you arrived at your \(K_m\) estimate. [3] (ii) Explain how the data in the table supports the conclusion that galactose acts as a competitive inhibitor of lactase. [3]
查看答案詳解收起答案詳解
解題
(a) To prepare 10 cm\(^3\) of each solution, use the formula \(C_1V_1 = C_2V_2\): For 2.0% lactose, mix 2.0 cm\(^3\) of the 10.0% stock with 8.0 cm\(^3\) of distilled water. For 4.0%, mix 4.0 cm\(^3\) stock with 6.0 cm\(^3\) water. For 6.0%, mix 6.0 cm\(^3\) stock with 4.0 cm\(^3\) water. For 8.0%, mix 8.0 cm\(^3\) stock with 2.0 cm\(^3\) water. For 10.0%, use 10.0 cm\(^3\) of the stock directly. Use a graduated pipette or syringe to measure these volumes. (b) Independent variable: substrate (lactose) concentration and the presence or absence of 1.0% galactose. Dependent variable: initial rate of reaction, measured by using a glucose biosensor or glucose reagent strip and a colorimeter to measure absorbance over time. Plot glucose concentration against time and calculate the gradient of the initial linear portion. Controlled variables: volume and concentration of enzyme (e.g., 2.0 cm\(^3\) of 2.0% lactase), temperature (using a water bath at 37 \(^{\circ}\)C), and pH (using a buffer of pH 6.5). Safety: Wear goggles and gloves to prevent skin or eye irritation from lactase. Repeat each run three times and calculate a mean. (c)(i) \(V_{\text{max}}\) is 50.0 a.u. Half of \(V_{\text{max}}\) is 25.0 a.u. By interpolation from the table without inhibitor, the lactose concentration corresponding to a rate of 25.0 a.u. is approximately 2.4% (accept any value from 2.2% to 2.5%). (c)(ii) At high substrate concentration (20.0%), the rate with the inhibitor (49.0 a.u.) is nearly equal to the maximum rate without inhibitor (50.0 a.u.), meaning \(V_{\text{max}}\) is unaffected. However, at lower substrate concentrations, the rate of reaction is significantly lower in the presence of the inhibitor, showing that competitive inhibition is overcome by high substrate concentrations.
評分準則
Part (a): [Max 3 marks] - 1 mark: Correctly identifies volumes of stock and water to make at least three concentrations. - 1 mark: Specifies using a syringe or graduated pipette for volume measurements. - 1 mark: Mentions stirring/mixing the prepared solutions. Part (b): [Max 6 marks] - 1 mark: Identifies independent variable (substrate concentration with and without inhibitor) and dependent variable (initial rate of glucose production). - 1 mark: Describes how to measure glucose concentration at regular intervals (e.g., every 30 seconds for 3 minutes) to determine the initial rate from the linear gradient. - 1 mark: Describes temperature control using a thermostatically controlled water bath. - 1 mark: Describes pH control using a buffer solution. - 1 mark: Describes enzyme concentration/volume control. - 1 mark: Mentions replication (minimum 3 repeats to calculate a mean) and a safety precaution (e.g., wearing gloves/goggles). Part (c)(i): [Max 3 marks] - 1 mark: Identifies \(V_{\text{max}}\) as 50.0 a.u. - 1 mark: Identifies \(K_m\) as the concentration at half of \(V_{\text{max}}\) (25.0 a.u.). - 1 mark: Estimates \(K_m\) as 2.2% to 2.5%. Part (c)(ii): [Max 3 marks] - 1 mark: Points out that at 20.0% substrate concentration, the rate with inhibitor reaches/approaches the maximum rate. - 1 mark: Explains that higher substrate concentration overcomes competitive inhibition by outcompeting the inhibitor for active sites. - 1 mark: Mentions that the inhibitor increases the apparent \(K_m\) without changing the maximum velocity of the enzyme.
題目 2 · Data Analysis & Methodology Design
15 分
A student investigated the effect of light intensity on the rate of photosynthesis of the water plant, *Elodea canadensis*. The student used a dissolved oxygen (DO) probe connected to a data logger to measure the oxygen concentration in a sealed chamber containing a 1.5 g sprig of *Elodea* immersed in a sodium hydrogencarbonate solution. A light source was placed at different distances from the chamber. (a) Identify three variables, other than the independent variable (light intensity), that must be controlled in this investigation and explain how each variable would be controlled. [3] (b) Explain why using a dissolved oxygen probe connected to a data logger is an improvement over counting bubbles of gas released by the plant. [3] (c) The student wants to convert the distance of the light source (\(d\), in cm) to relative light intensity (\(I\)) using the inverse square law: \(I = \frac{1}{d^2}\). (i) Write down the formula relating relative light intensity (\(I\)) to distance (\(d\)). [1] (ii) Calculate the relative light intensity at a distance of 40 cm. Show your working and state the units. [1] (d) The student obtained the following results at a distance of 20 cm: Initial oxygen concentration = 6.2 mg dm\(^{-3}\); Final oxygen concentration after 10 minutes = 8.8 mg dm\(^{-3}\); Volume of water in the chamber = 0.15 dm\(^3\). Calculate the rate of oxygen production in mg g\(^{-1}\) hour\(^{-1}\). Show your working. [3] (e) To analyze whether there is a significant difference between the rates of photosynthesis measured at a distance of 20 cm and at 40 cm, the student decided to perform a statistical test. (i) State the name of a suitable statistical test for this analysis and justify your choice. [2] (ii) State a suitable null hypothesis for this statistical test. [2]
查看答案詳解收起答案詳解
解題
(a) 1. Temperature: Controlled by placing a glass heat-absorbing screen or water jacket between the light source and the chamber, or using an LED bulb that emits very little heat. 2. Carbon dioxide concentration: Controlled by using a fixed volume and concentration of sodium hydrogencarbonate solution (e.g., 0.1 mol dm\(^{-3}\) solution). 3. Plant characteristics: Use the exact same sprig of *Elodea* for all experimental runs, or select sprigs with matching mass and number of leaves. (b) 1. Measuring oxygen concentration directly with a DO probe is quantitative, avoiding human error associated with counting bubbles. 2. Bubbles can vary in size and rate, making bubble counting inaccurate, while the probe measures actual dissolved oxygen mass. 3. The data logger records continuous, real-time values with high precision and resolution. (c)(i) \(I = \frac{1}{d^2}\) (c)(ii) \(I = \frac{1}{40^2} = \frac{1}{1600} = 0.000625\) arbitrary units (or \(6.25 \times 10^{-4}\) a.u.). (d) Step 1: Net increase in oxygen concentration = \(8.8 - 6.2 = 2.6\) mg dm\(^{-3}\). Step 2: Total mass of oxygen produced = \(2.6 \text{ mg dm}^{-3} \times 0.15 \text{ dm}^3 = 0.39\) mg. Step 3: Convert time to hours: \(10 \text{ minutes} = \frac{10}{60} = 0.1667\) hours. Step 4: Scale for plant mass and time: \(\text{Rate} = \frac{0.39 \text{ mg}}{1.5 \text{ g} \times 0.1667 \text{ hours}} = 1.56\) mg g\(^{-1}\) hour\(^{-1}\). (e)(i) Student's t-test (or t-test). Justification: It compares the means of two sets of continuous, normally distributed data. (e)(ii) There is no significant difference between the mean rate of oxygen production (photosynthesis) of *Elodea* at a light source distance of 20 cm and at 40 cm.
評分準則
Part (a): [Max 3 marks, 1 per variable + method] - 1 mark: Temperature control (e.g., LED lamp or water screen). - 1 mark: CO2 concentration control (e.g., fixed NaHCO\(_3\) concentration). - 1 mark: Plant state control (e.g., using same sprig / identical mass and length). Part (b): [Max 3 marks] - 1 mark: Bubble size is variable / bubbles can be missed by human observers. - 1 mark: DO probe measures direct mass/concentration of oxygen, not an indirect estimate. - 1 mark: Data logger collects continuous, objective, high-resolution data automatically. Part (c)(i): [1 mark] - Correctly states the formula: \(I = \frac{1}{d^2}\). Part (c)(ii): [1 mark] - Correctly calculates \(0.000625\) (or \(6.25 \times 10^{-4}\)) arbitrary units. Part (d): [Max 3 marks] - 1 mark: Calculates total mass of O2 produced: \(2.6 \times 0.15 = 0.39\) mg. - 1 mark: Scales to hourly rate: \(0.39 \times 6 = 2.34\) mg hour\(^{-1}\). - 1 mark: Scales to mass of plant: \(1.56\) mg g\(^{-1}\) hour\(^{-1}\). Part (e)(i): [Max 2 marks] - 1 mark: Identifies Student's t-test. - 1 mark: Justifies because it compares the means of two independent groups/treatments of continuous data. Part (e)(ii): [2 marks] - 2 marks: Formulates a complete null hypothesis stating no significant difference between the mean rate of photosynthesis at 20 cm and 40 cm.
想知道自己有幾分把握？
Thinka 是 DSE 學生用的 AI 練習應用程式，有無限量練習題、即時自動批改和詳細解題步驟。逾 100,000 名學生用它確認自己真的識，而不只是「以為識」。