An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 SL (TZ1) IB Diploma Programme Biology paper. Not affiliated with or reproduced from IB.
Paper 1A
Answer all 30 multiple-choice questions. A calculator is required.
30 題目 · 30 分
題目 1 · 選擇題
1 分
An ecologist wants to estimate the population size of a forest beetle species using the mark-release-recapture method. On the first day, they capture, mark, and release 80 beetles. One week later, they capture 100 beetles, of which 20 are found to have been marked. What is the estimated population size of the beetles, and which option describes a key assumption of this method?
A.400; Marked individuals mix randomly with the unmarked population.
B.400; Emigration rates from the woodland are significantly higher than immigration rates.
C.1600; No births or deaths occur during the sampling period.
D.1600; Marked individuals have a higher likelihood of being recaptured due to trap-happy behavior.
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解題
The Lincoln Index formula is defined as \(N = \frac{n_1 \times n_2}{m}\), where \(n_1\) is the number of individuals caught and marked in the first sample (80), \(n_2\) is the number of individuals caught in the second sample (100), and \(m\) is the number of marked individuals recaptured (20). Calculating this gives: \(N = \frac{80 \times 100}{20} = 400\). A key assumption of this method is that marked individuals mix randomly with the rest of the unmarked population before the second capture occurs.
評分準則
Award 1 mark for the correct calculation of 400 and the correct identification of the random mixing assumption.
題目 2 · 選擇題
1 分
Which of the following describes the correct sequence of events that occurs in the human body in response to a rise in blood solute concentration?
A.Osmoreceptors in the hypothalamus detect the change \(\rightarrow\) pituitary gland releases more ADH \(\rightarrow\) aquaporins are inserted into the collecting duct membranes \(\rightarrow\) water reabsorption increases.
B.Osmoreceptors in the medulla oblongata detect the change \(\rightarrow\) pituitary gland releases less ADH \(\rightarrow\) aquaporins are retrieved from the collecting duct membranes \(\rightarrow\) water reabsorption decreases.
C.Osmoreceptors in the hypothalamus detect the change \(\rightarrow\) adrenal gland releases aldosterone \(\rightarrow\) active transport of sodium ions out of the collecting duct \(\rightarrow\) water reabsorption decreases.
D.Osmoreceptors in the glomerulus detect the change \(\rightarrow\) pituitary gland releases more ADH \(\rightarrow\) collecting duct becomes less permeable to water \(\rightarrow\) urine volume increases.
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解題
When blood solute concentration increases (dehydration), osmoreceptors in the hypothalamus detect the change and stimulate the posterior pituitary gland to secrete more antidiuretic hormone (ADH). ADH travels in the blood to the kidneys, where it binds to receptors on the collecting duct, triggering the insertion of aquaporins into the luminal membrane. This increases the water permeability of the collecting duct, leading to greater water reabsorption back into the blood.
評分準則
Award 1 mark for the correct physiological sequence beginning with hypothalamic osmoreceptors, leading to ADH release, aquaporin insertion, and increased water reabsorption.
題目 3 · 選擇題
1 分
During aerobic cell respiration in eukaryotes, where does substrate-level phosphorylation take place, and how many net molecules of ATP are produced by this pathway per molecule of glucose?
A.Cytoplasm and mitochondrial matrix; 4 ATP
B.Mitochondrial matrix only; 2 ATP
C.Inner mitochondrial membrane; 32 to 34 ATP
D.Cytoplasm and inner mitochondrial membrane; 36 ATP
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解題
Substrate-level phosphorylation occurs during glycolysis in the cytoplasm (producing a net of 2 ATP) and during the Krebs cycle in the mitochondrial matrix (producing 1 ATP per turn of the cycle, which equals 2 ATP per molecule of glucose). Thus, substrate-level phosphorylation occurs in both the cytoplasm and mitochondrial matrix, yielding a total of 4 net ATP molecules per glucose molecule. The remaining ATP is produced via oxidative phosphorylation at the inner mitochondrial membrane.
評分準則
Award 1 mark for correctly identifying both cellular locations and the total count of 4 net ATP molecules.
題目 4 · 選擇題
1 分
An enzyme-catalyzed reaction is monitored in the presence of an inhibitor. It is observed that increasing the substrate concentration to a very high level fully restores the maximum rate of reaction (\(V_{\max}\)) to the level observed without the inhibitor. What type of inhibitor is this, and what is its effect on the Michaelis constant (\(K_m\)) of the enzyme?
A.Competitive inhibitor; increases the \(K_m\)
B.Non-competitive inhibitor; decreases the \(K_m\)
C.Competitive inhibitor; decreases the \(K_m\)
D.Non-competitive inhibitor; does not change the \(K_m\)
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解題
A competitive inhibitor competes with the substrate for binding at the active site. Because high substrate concentrations can outcompete the inhibitor, the maximum velocity (\(V_{\max}\)) is unchanged. However, because it requires a higher concentration of substrate to achieve half of this maximum velocity, the Michaelis constant (\(K_m\)) is increased (reflecting a decreased apparent affinity of the enzyme for its substrate).
評分準則
Award 1 mark for identifying the inhibitor as competitive and stating that the Michaelis constant (\(K_m\)) increases.
題目 5 · 選擇題
1 分
During the repolarization phase of an action potential in a human neuron, what is the state of the voltage-gated ion channels and the direction of net ion movement?
A.Sodium channels are inactivated; potassium channels are open, and \(K^+\) ions diffuse out of the axon.
B.Sodium channels are open; potassium channels are closed, and \(Na^+\) ions diffuse into the axon.
C.Sodium channels are closed; potassium channels are open, and \(K^+\) ions are actively pumped into the axon.
D.Sodium channels are inactivated; potassium channels are inactivated, and both ions are restored by the sodium-potassium pump.
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解題
During the repolarization phase of an action potential, voltage-gated sodium (\(Na^+\)) channels close and become inactivated, halting the influx of sodium. Simultaneously, voltage-gated potassium (\(K^+\)) channels open, allowing potassium ions to diffuse rapidly out of the axon down their electrochemical gradient. This efflux of positive charge restores the negative resting membrane potential.
評分準則
Award 1 mark for the correct channel states (sodium inactivated, potassium open) and the correct movement of potassium ions diffusing out of the cell.
題目 6 · 選擇題
1 分
In a plant species, the gene for red flowers (\(R\)) is dominant over white flowers (\(r\)), and the gene for tall stems (\(T\)) is dominant over short stems (\(t\)). A heterozygous plant (\(RrTt\)) with the dominant alleles on one chromosome and recessive alleles on the homologous chromosome (coupling/cis linkage, represented as \(\frac{RT}{rt}\)) is crossed with a double homozygous recessive plant (\(\frac{rt}{rt}\)). If the distance between the two gene loci is 15 centimorgans (cM), what is the expected percentage of offspring with the red flower, short stem phenotype?
A.7.5%
B.15%
C.42.5%
D.25%
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解題
The parental gametes produced by the heterozygous plant are \(RT\) and \(rt\). The recombinant gametes are \(Rt\) and \(rT\). The map distance of 15 cM corresponds to a recombination frequency of 15% in total. Because there are two recombinant classes (\(Rt\) and \(rT\)), each individual recombinant phenotype is expected to occur with equal frequency. Therefore, the frequency of the red-flowered, short-stemmed recombinant offspring (\(\frac{Rt}{rt}\)) is \(\frac{15\%}{2} = 7.5\%\).
評分準則
Award 1 mark for calculating the correct percentage of 7.5% based on the recombination frequency divided by two.
題目 7 · 選擇題
1 分
A transfer RNA (tRNA) molecule carries the amino acid methionine and possesses the anticodon sequence 5'-CAU-3'. Which codon on the messenger RNA (mRNA) will this tRNA molecule pair with during translation, and what type of bond is formed when methionine is linked to the growing polypeptide chain?
A.5'-AUG-3'; peptide bond
B.5'-GUA-3'; peptide bond
C.5'-AUG-3'; hydrogen bond
D.5'-UAC-3'; ester bond
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解題
Codon-anticodon base pairing is complementary and antiparallel. For an anticodon running 5'-CAU-3', the complementary mRNA codon must run 3'-GUA-5', which is written in the standard 5' to 3' direction as 5'-AUG-3'. During translation, amino acids are joined together by covalent peptide bonds.
評分準則
Award 1 mark for the correct codon sequence of 5'-AUG-3' paired with the identification of a peptide bond.
題目 8 · 選擇題
1 分
An ecology student uses 100 quadrats to investigate the distribution of two saltmarsh plant species: *Salicornia* and *Spartina*. *Salicornia* is found in 60 of the quadrats, and *Spartina* is found in 50 of the quadrats. If the distribution of these two species is completely independent, what is the expected frequency of quadrats containing both species, and what statistical test should be used to analyze the actual association?
A.30; Chi-squared test of independence
B.40; Student's t-test
C.30; Spearman's rank correlation
D.10; Chi-squared goodness-of-fit test
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解題
If the distributions of the two species are independent, the probability of both occurring in the same quadrat is the product of their individual probabilities. Probability of *Salicornia* = \(\frac{60}{100} = 0.6\); Probability of *Spartina* = \(\frac{50}{100} = 0.5\). The joint probability is \(0.6 \times 0.5 = 0.3\). In a total sample of 100 quadrats, the expected frequency is \(0.3 \times 100 = 30\) quadrats. The appropriate statistical test to analyze association from contingency table data is the Chi-squared test of independence.
評分準則
Award 1 mark for the correct calculation of the expected frequency (30) and the correct identification of the Chi-squared test of independence.
題目 9 · 選擇題
1 分
In an ecosystem, Species X is a dominant plant competitor that excludes other species. If a selective herbivore is introduced that feeds exclusively on Species X, what is the expected long-term effect on the carrying capacity \(K\) of the other competing plant species and on the overall species richness of the community?
A.Carrying capacity \(K\) of competitors increases, and species richness increases.
B.Carrying capacity \(K\) of competitors increases, but species richness decreases.
C.Carrying capacity \(K\) of competitors decreases, and species richness increases.
D.Carrying capacity \(K\) of competitors remains unchanged, but species richness decreases.
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解題
Reducing the population of the dominant competitor (Species X) relieves competitive exclusion. This allows competing species to utilize shared resources more effectively, increasing their carrying capacity \(K\). Furthermore, preventing one species from dominating allows more species to coexist, which increases the overall species richness of the community.
評分準則
Award 1 mark for the correct answer (A). Award 0 marks for incorrect options.
題目 10 · 選擇題
1 分
During exposure to cold conditions, the human body initiates physiological mechanisms to maintain homeostatic core temperature. Which of the following correctly pairs the physiological response, its immediate effector tissue, and the type of feedback loop involved?
A.Vasoconstriction of skin arterioles | Smooth muscle in arteriole walls | Negative feedback
In cold conditions, negative feedback pathways trigger the contraction of smooth muscle in the walls of skin arterioles, causing vasoconstriction. This reduces blood flow to the skin surface to conserve heat. Shivering is also a negative feedback mechanism, not positive feedback.
評分準則
Award 1 mark for the correct answer (A). Award 0 marks for incorrect options.
題目 11 · 選擇題
1 分
A suspension of intact, functional mitochondria is incubated in a buffered solution containing ADP and inorganic phosphate. Which substrate must be added to the suspension to initiate a high rate of electron transport and oxygen consumption?
A.Glucose
B.Pyruvate
C.Fructose-1,6-bisphosphate
D.Lactate
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解題
Intact mitochondria lack the enzymes necessary for glycolysis, which occur in the cytoplasm. Therefore, they cannot directly metabolize glucose or fructose-1,6-bisphosphate. Pyruvate, however, can be transported directly into the mitochondrial matrix, where it is converted to acetyl-CoA to fuel the Krebs cycle and the electron transport chain, causing oxygen consumption.
評分準則
Award 1 mark for the correct answer (B). Award 0 marks for incorrect options.
題目 12 · 選擇題
1 分
An enzyme-catalyzed reaction is studied in the presence of a metabolic inhibitor. When the substrate concentration is increased to very high levels, the rate of reaction is restored to the maximum velocity \(V_{max}\) observed in the absence of the inhibitor. Which statement correctly identifies the type of inhibition and its effect on the Michaelis constant \(K_m\)?
A.Competitive inhibition; the \(K_m\) increases
B.Non-competitive inhibition; the \(K_m\) remains unchanged
C.Competitive inhibition; the \(K_m\) decreases
D.Non-competitive inhibition; the \(K_m\) increases
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解題
If \(V_{max}\) can be restored at high substrate concentrations, the inhibitor must be competitive, as the substrate can outcompete the inhibitor for binding to the active site. Because a higher concentration of substrate is required to reach half of \(V_{max}\), the apparent Michaelis constant \(K_m\) increases.
評分準則
Award 1 mark for the correct answer (A). Award 0 marks for incorrect options.
題目 13 · 選擇題
1 分
How does the myelin sheath increase the speed of conduction of an action potential along an axon?
A.It decreases the electrical resistance of the axoplasm, allowing ions to flow faster.
B.It permits continuous depolarization to occur along the entire length of the membrane.
C.It acts as an electrical insulator, restricting depolarization and ion exchange to the nodes of Ranvier.
D.It increases the density of voltage-gated sodium channels under the insulated sheath.
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解題
Myelin acts as an electrical insulator, preventing ion flow across the axon membrane. Consequently, action potentials can only occur at the unmyelinated gaps, known as the nodes of Ranvier. This causes the action potential to jump from node to node, a process called saltatory conduction, which greatly increases transmission speed.
評分準則
Award 1 mark for the correct answer (C). Award 0 marks for incorrect options.
題目 14 · 選擇題
1 分
In fruit flies, eye color is determined by a gene on the X chromosome. The allele for red eyes \(X^R\) is dominant to the allele for white eyes \(X^r\). If a white-eyed female fly is crossed with a red-eyed male fly, what are the expected phenotypes of the \(F_1\) generation offspring?
A.All female offspring will have red eyes, and all male offspring will have white eyes.
B.All female offspring will have white eyes, and all male offspring will have red eyes.
C.All offspring, both male and female, will have red eyes.
D.Half of the females will have red eyes, half of the females will have white eyes, and all males will have white eyes.
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解題
The white-eyed female has the genotype \(X^r X^r\). The red-eyed male has the genotype \(X^R Y\). Female offspring receive an \(X^r\) from the mother and an \(X^R\) from the father, resulting in the heterozygous genotype \(X^R X^r\) (red eyes). Male offspring receive an \(X^r\) from the mother and a \(Y\) from the father, resulting in the genotype \(X^r Y\) (white eyes). Therefore, all females have red eyes and all males have white eyes.
評分準則
Award 1 mark for the correct answer (A). Award 0 marks for incorrect options.
題目 15 · 選擇題
1 分
During the elongation phase of translation in eukaryotes, which event occurs immediately after a peptide bond is formed between the amino acid in the A site and the growing polypeptide chain in the P site?
A.The small ribosomal subunit detaches from the mRNA template.
B.The ribosome translocates three nucleotides along the mRNA in the \(5'\) to \(3'\) direction.
C.An aminoacyl-tRNA binds directly to the E site of the ribosome.
D.The uncharged tRNA in the A site is immediately released into the cytoplasm.
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解題
Following peptide bond formation, translocation occurs. The ribosome moves three nucleotides (one codon) along the mRNA in the \(5'\) to \(3'\) direction. This shifts the peptidyl-tRNA from the A site to the P site, and the uncharged tRNA from the P site to the E site, allowing elongation to continue.
評分準則
Award 1 mark for the correct answer (B). Award 0 marks for incorrect options.
題目 16 · 選擇題
1 分
After a meal high in carbohydrates, blood glucose concentration rises. Which response correctly describes how the pancreas helps restore homeostatic blood glucose levels?
A.Alpha cells secrete glucagon, which stimulates glycogenesis in liver cells.
B.Beta cells secrete insulin, which stimulates glycogenolysis in liver cells.
C.Beta cells secrete insulin, which promotes glucose uptake by body cells and glycogenesis in the liver.
D.Alpha cells secrete insulin, which decreases the rate of cellular respiration in muscle cells.
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解題
Elevated blood glucose is detected by the beta cells of the pancreas, which secrete insulin. Insulin lowers blood glucose levels by promoting glucose uptake into body cells (e.g., muscle and fat cells) and stimulating glycogenesis (the synthesis of glycogen from glucose) in the liver and muscle tissues.
評分準則
Award 1 mark for the correct answer (C). Award 0 marks for incorrect options.
題目 17 · 選擇題
1 分
A student uses the Lincoln index to estimate the population size of a beetle species in a local woodland. On Monday, they capture, mark, and release 120 beetles. On Tuesday, they capture 150 beetles. Assuming the true population size is 600, but 20% of the marks washed off between Monday and Tuesday, what is the expected calculated estimate of the population size?
A.750, which is an overestimate of the true population.
B.750, which is an underestimate of the true population.
C.480, which is an underestimate of the true population.
D.480, which is an overestimate of the true population.
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解題
1. Under normal circumstances with a true population of 600, the proportion of marked beetles is \( \frac{120}{600} = 0.20 \). 2. If 20% of the marks wash off, only 96 beetles remain marked: \( 120 \times 0.80 = 96 \). 3. The true proportion of marked beetles in the population becomes \( \frac{96}{600} = 0.16 \). 4. In the second sample of 150 beetles, the expected number of marked beetles recaptured is \( 150 \times 0.16 = 24 \). 5. The student, unaware of the mark loss, calculates the estimated population size using the standard formula: \( N = \frac{n_1 \times n_2}{m_2} = \frac{120 \times 150}{24} = 750 \). 6. Since the true population is 600 and the estimate is 750, this is an overestimate.
評分準則
Award 1 mark for selecting the correct option (A). - Reject other options because they represent incorrect calculations or incorrect classifications of the estimate.
題目 18 · 選擇題
1 分
A study was conducted to determine if there is an association between two plant species, *Plantago major* and *Taraxacum officinale*, in a meadow. Out of 100 quadrats sampled, both species were present in 30 quadrats, only *P. major* was in 20, only *T. officinale* was in 10, and neither was in 40. Under the null hypothesis of independence, what is the expected frequency of quadrats containing both species, and what type of association is suggested?
A.Expected frequency is 20, suggesting a positive association.
B.Expected frequency is 20, suggesting a negative association.
C.Expected frequency is 30, suggesting no association.
D.Expected frequency is 12, suggesting a positive association.
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解題
1. Determine total presence of each species: - *P. major* present in: \( 30 + 20 = 50 \) quadrats. - *T. officinale* present in: \( 30 + 10 = 40 \) quadrats. 2. Total number of quadrats sampled = 100. 3. Expected frequency of quadrats with both species under independence: \( \frac{50 \times 40}{100} = 20 \). 4. Since the observed frequency of both species being present (30) is greater than the expected frequency (20), this suggests a positive association.
評分準則
Award 1 mark for the correct option (A). - Reject B because it suggests a negative association. - Reject C and D because the expected frequency calculation is incorrect.
題目 19 · 選擇題
1 分
A patient suffers from an autoimmune disease that selectively destroys their pancreatic beta (\(\beta\)) cells, while leaving their alpha (\(\alpha\)) cells completely intact. Which of the following describes the most likely physiological status of this patient two hours after consuming a high-carbohydrate meal compared to a healthy individual?
A.Blood insulin remains low, glucagon remains high, and liver glycogen synthesis is decreased.
B.Blood insulin is high, blood glucagon is low, and liver glycogenolysis is increased.
C.Blood insulin remains low, blood glucagon is low, and liver gluconeogenesis is decreased.
D.Blood insulin is high, glucagon remains high, and liver glycogen synthesis is increased.
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解題
1. Destruction of pancreatic beta cells prevents the secretion of insulin in response to high blood glucose. 2. In a healthy individual, insulin inhibits glucagon secretion from alpha cells. Without insulin, glucagon secretion remains inappropriately high. 3. Because insulin levels are low, insulin-dependent pathways such as glycogen synthesis (glycogenesis) in the liver are severely decreased.
評分準則
Award 1 mark for the correct choice (A). - Reject B and D because insulin levels cannot be high without functional beta cells. - Reject C because glucagon remains high, not low, due to the loss of insulin's inhibitory effect.
題目 20 · 選擇題
1 分
An experimental drug acts as an inhibitor of cellular respiration by blocking the transfer of electrons from Complex III to cytochrome c within the electron transport chain. What are the immediate consequences of adding this inhibitor to active mitochondria?
A.Oxygen consumption increases, and the proton gradient across the inner membrane increases.
B.Oxygen consumption decreases, and ATP synthesis decreases.
C.The Krebs cycle rate increases to compensate, and NADH concentration decreases.
D.Carbon dioxide production in the link reaction ceases immediately.
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解題
1. If electrons cannot be transferred from Complex III to cytochrome c, the downstream flow of electrons to oxygen (via Complex IV) is blocked. 2. This prevents oxygen from acting as the terminal electron acceptor, leading to a decrease in oxygen consumption. 3. The proton-pumping activity of the electron transport chain slows down, reducing the proton gradient across the inner mitochondrial membrane and consequently decreasing ATP synthesis.
評分準則
Award 1 mark for the correct option (B). - Reject A because oxygen consumption and the proton gradient would decrease, not increase. - Reject C because the Krebs cycle rate would slow down due to NADH accumulation. - Reject D because the link reaction occurs in the matrix and is not immediately coupled to ETC blocking in a way that stops carbon dioxide production instantaneously.
題目 21 · 選擇題
1 分
Consider the metabolic pathway: \( A \xrightarrow{E_1} B \xrightarrow{E_2} C \xrightarrow{E_3} D \). Under normal conditions, metabolite D acts as an allosteric inhibitor of enzyme \(E_1\). If a researcher introduces a non-competitive inhibitor of enzyme \(E_3\), how will the concentrations of metabolites B, C, and D change?
A.B increases, C increases, and D decreases.
B.B decreases, C increases, and D increases.
C.B increases, C decreases, and D decreases.
D.B decreases, C decreases, and D increases.
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解題
1. Inhibiting \(E_3\) blocks the conversion of C to D, resulting in an accumulation of C and a depletion of D. 2. Since the concentration of D decreases, the allosteric feedback inhibition of D on enzyme \(E_1\) is relieved. 3. With \(E_1\) highly active and no longer regulated by feedback inhibition, the rate of conversion of A to B increases, leading to an increase in the concentration of B (as well as C, which cannot proceed past \(E_3\)).
評分準則
Award 1 mark for the correct option (A). - Reject other options as they incorrectly predict the accumulation or depletion trends of the intermediates based on the feedback loop mechanisms.
題目 22 · 選擇題
1 分
A neurotoxin extracted from a marine snail selectively blocks voltage-gated potassium channels in axonal membranes. How would this toxin affect the shape of an action potential recording?
A.The membrane fails to reach the threshold potential.
B.The depolarization phase is much faster and reaches a higher peak.
C.The repolarization phase is prolonged, extending the duration of the action potential.
D.The resting membrane potential immediately becomes more negative.
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解題
1. Depolarization relies on voltage-gated sodium channels opening, allowing sodium ions to enter the axon. This phase remains unaffected by the toxin. 2. Repolarization relies on the opening of voltage-gated potassium channels, allowing potassium ions to leave the axon to restore the resting membrane potential. 3. If these potassium channels are blocked, repolarization is severely delayed (relying only on passive leak channels), which prolongs the duration of the action potential.
評分準則
Award 1 mark for the correct option (C). - Reject A because the threshold potential is not determined by voltage-gated potassium channels. - Reject B because depolarization kinetics depend on sodium channels. - Reject D because resting membrane potential is maintained primarily by leak channels and the sodium-potassium pump, not voltage-gated channels.
題目 23 · 選擇題
1 分
In sweet pea plants, the allele for purple flowers (P) is dominant over red flowers (p), and the allele for long pollen grains (L) is dominant over round pollen grains (l). A plant heterozygous for both traits (PpLl) is test-crossed with a double recessive plant (ppll). The resulting offspring have the following phenotypes: Purple, long (42%); Red, round (44%); Purple, round (7%); Red, long (7%). What can be concluded from these results?
A.The genes for flower colour and pollen shape are on different chromosomes and assort independently.
B.The genes are linked, and the distance between them is approximately 14 map units.
C.The genes are linked, and the recombinant frequency is 86%.
D.The flower colour gene is sex-linked, while the pollen shape gene is autosomal.
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解題
1. A test cross with independent assortment yields a 1:1:1:1 ratio (25% each). 2. Here, the parental phenotypes (Purple/long and Red/round) are highly overrepresented, indicating gene linkage. 3. The recombinant phenotypes are Purple/round (7%) and Red/long (7%). 4. Recombination frequency = \( 7\% + 7\% = 14\% \). 5. A 14% recombination frequency corresponds to a genetic distance of approximately 14 map units (centimorgans).
評分準則
Award 1 mark for selecting the correct option (B). - Reject A because the offspring ratio deviates significantly from independent assortment. - Reject C because 86% is the parental frequency, not the recombinant frequency. - Reject D because there is no evidence of sex-linked inheritance patterns in this classical autosomal linkage example.
題目 24 · 選擇題
1 分
During translation, a codon on the mRNA molecule has the sequence 5'-AUG-3'. What is the corresponding sequence of the anticodon on the tRNA molecule (written in the 5' to 3' direction), and to which end of the tRNA molecule is the amino acid attached?
A.Anticodon is 5'-CAU-3' and the amino acid is attached to the 3' end.
B.Anticodon is 5'-UAC-3' and the amino acid is attached to the 5' end.
C.Anticodon is 5'-CAU-3' and the amino acid is attached to the 5' end.
D.Anticodon is 5'-UAC-3' and the amino acid is attached to the 3' end.
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解題
1. The mRNA codon is 5'-AUG-3'. 2. The complementary antiparallel tRNA anticodon sequence is 3'-UAC-5'. 3. Written in the standard 5' to 3' direction, this sequence is 5'-CAU-3'. 4. Amino acids are always covalently attached to the CCA terminal sequence at the 3' end of the tRNA molecule.
評分準則
Award 1 mark for the correct option (A). - Reject options with incorrect anticodon directions (such as 5'-UAC-3') or incorrect amino acid attachment ends (5' end).
題目 25 · 選擇題
1 分
An ecologist uses the Lincoln index (mark-release-recapture) to estimate the population size of a species of beetle in a local grassland. Which of the following occurrences during the study period would result in an overestimation of the true population size?
A.A significant proportion of the marked beetles lose their non-toxic paint marks before the second sampling.
B.The marked beetles become habituated to the traps and are more likely to be caught again.
C.Unmarked beetles migrate out of the study area before the second sampling.
D.A small number of unmarked beetles die of natural causes during the week between samples.
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解題
The Lincoln index formula is \(N = \frac{n_1 \times n_2}{m_2}\), where \(n_1\) is the number of individuals marked in the first sample, \(n_2\) is the total number of individuals caught in the second sample, and \(m_2\) is the number of marked individuals recaptured. If marked beetles lose their marks, they will be counted as unmarked in the second sample, decreasing the value of \(m_2\). A smaller denominator \(m_2\) results in a larger calculated population size \(N\), leading to an overestimation.
評分準則
Award 1 mark for the correct option (A). Reject other options because trap-habituation (B) increases recaptures (underestimation), and loss of unmarked individuals (C or D) does not directly decrease the proportion of marked individuals recaptured in the same way.
題目 26 · 選擇題
1 分
Dehydration causes a change in blood osmolarity, triggering a homeostatic response to conserve water. Which row in the table below correctly describes the initial changes in blood solute concentration, the secretion of antidiuretic hormone (ADH), and the water permeability of the kidney collecting duct cells?
Dehydration means loss of water, which increases the concentration of solutes in the blood. This high solute concentration is detected by osmoreceptors in the hypothalamus, which stimulates the pituitary gland to secrete more ADH. ADH acts on the cells of the collecting duct to make them more permeable to water (via insertion of aquaporins), allowing more water to be reabsorbed into the blood.
評分準則
Award 1 mark for the correct choice A. Choice B, C, and D contain incorrect relationships between dehydration, hormone release, and target organ response.
題目 27 · 選擇題
1 分
Mitochondria are treated with a metabolic poison that binds to and blocks cytochrome c oxidase (Complex IV) of the electron transport chain. What would be the immediate effect of this poison on the proton gradient across the inner mitochondrial membrane and the rate of ATP synthesis via chemiosmosis?
A.Proton gradient: Decreases | ATP synthesis rate: Decreases
B.Proton gradient: Increases | ATP synthesis rate: Decreases
C.Proton gradient: Decreases | ATP synthesis rate: Increases
D.Proton gradient: Unchanged | ATP synthesis rate: Unchanged
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解題
When Complex IV is blocked, electrons can no longer be transferred to oxygen (the final electron acceptor). This halts the movement of electrons along the entire electron transport chain. As a result, protons are no longer pumped from the mitochondrial matrix into the intermembrane space. The existing proton gradient will dissipate as protons flow back through ATP synthase without being replenished, leading to a decrease in the proton gradient and a subsequent decrease in the rate of ATP synthesis.
評分準則
Award 1 mark for the correct choice A. B, C, and D are incorrect because blocking electron transport prevents proton pumping, reducing the gradient and halting ATP synthesis.
題目 28 · 選擇題
1 分
How do competitive and non-competitive inhibitors affect the maximum rate of reaction (\(V_{max}\)) and the Michaelis constant (\(K_m\), the substrate concentration at which the reaction rate is half of \(V_{max}\)) of an enzyme?
A.Competitive: \(V_{max}\) is unchanged, \(K_m\) increases | Non-competitive: \(V_{max}\) decreases, \(K_m\) is unchanged
B.Competitive: \(V_{max}\) decreases, \(K_m\) is unchanged | Non-competitive: \(V_{max}\) is unchanged, \(K_m\) increases
Competitive inhibitors compete with the substrate for the active site of the enzyme. At very high substrate concentrations, the substrate outcompetes the inhibitor, allowing the enzyme to reach its original maximum rate of reaction (\(V_{max}\) is unchanged), but a higher concentration of substrate is needed to reach half of \(V_{max}\) (\(K_m\) increases). Non-competitive inhibitors bind to an allosteric site, disabling the catalytic activity of the enzyme regardless of substrate concentration (\(V_{max}\) decreases), but the affinity of the unaffected enzymes remains the same (\(K_m\) is unchanged).
評分準則
Award 1 mark for the correct choice A. B, C, and D represent incorrect combinations of the effects of inhibitors on \(V_{max}\) and \(K_m\).
題目 29 · 選擇題
1 分
Which of the following events is directly responsible for the rapid repolarization phase of an action potential in a mammalian neuron?
A.The active transport of sodium ions out of the axon by the sodium-potassium pump.
B.The diffusion of potassium ions out of the axon through voltage-gated potassium channels.
C.The diffusion of sodium ions into the axon through voltage-gated sodium channels.
D.The closing of non-gated potassium leak channels.
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解題
Repolarization is the phase of the action potential where the membrane potential returns to a negative value. This is achieved by the opening of voltage-gated potassium channels, which allows potassium ions (\(K^+\)) to rapidly diffuse down their electrochemical gradient out of the axon. This loss of positive charge restores the negative internal charge of the neuron.
評分準則
Award 1 mark for the correct answer B. A is incorrect because the sodium-potassium pump restores long-term ion gradients, not the rapid electrical changes of repolarization. C describes depolarization. D would prevent repolarization.
題目 30 · 選擇題
1 分
A test cross is performed between a female fruit fly heterozygous for two autosomal traits (grey body, normal wings: \(GgWw\)) and a male fly with a black body and vestigial wings (\(ggww\)). The offspring phenotypes and counts are: Grey body, normal wings: 415; Black body, vestigial wings: 405; Grey body, vestigial wings: 92; Black body, normal wings: 88. What is the recombinant frequency, and what does this indicate about the loci of the genes?
A.Recombinant frequency is 9%; the genes are on different chromosomes (unlinked).
B.Recombinant frequency is 18%; the genes are on the same chromosome (linked).
C.Recombinant frequency is 18%; the genes are on different chromosomes (unlinked).
D.Recombinant frequency is 82%; the genes are on the same chromosome (linked).
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解題
Recombinants are the offspring with non-parental phenotypes (Grey, vestigial and Black, normal). Total offspring = \(415 + 405 + 92 + 88 = 1000\). Total recombinants = \(92 + 88 = 180\). Recombinant frequency = \(\frac{180}{1000} \times 100\% = 18\%\). Since the recombinant frequency is significantly less than 50%, the genes do not assort independently, indicating that they are linked on the same chromosome.
評分準則
Award 1 mark for the correct calculation and interpretation (B). A and D have incorrect calculations. C has the correct calculation but incorrect interpretation of linkage (recombinant frequency < 50% indicates linkage).
Paper 1B
Answer all short-answer data-based questions in the boxes provided.
4 題目 · 25 分
題目 1 · Data-based & Short-answer
6.25 分
In a study on the interactions of two desert plant species, Agave americana and Larrea tridentata, researchers measured the root density (g dm^-3) at various distances from the stem of Larrea tridentata when grown in monoculture versus in mixed culture with Agave americana.
Data Table: - Distance from stem (m): 0.5, 1.0, 1.5, 2.0 - Monoculture root density of L. tridentata (g dm^-3): 12.0, 8.0, 4.0, 1.2 - Mixed culture root density of L. tridentata (g dm^-3): 5.0, 3.0, 1.0, 0.2
Questions: (a) Calculate the percentage decrease in root density of L. tridentata at a distance of 1.0 m when grown in mixed culture compared to monoculture. [2] (b) Describe the effect of the presence of Agave americana on the root density of Larrea tridentata across all distances. [2] (c) Explain how these results provide evidence for interspecific competition between the two species. [2]
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解題
Part a: Formula for percentage decrease = ((Monoculture value - Mixed culture value) / Monoculture value) * 100. At 1.0 m, Monoculture = 8.0 g dm^-3, Mixed culture = 3.0 g dm^-3. Difference = 8.0 - 3.0 = 5.0 g dm^-3. Percentage decrease = (5.0 / 8.0) * 100 = 62.5%. Part b: Root density is lower at every distance measured in the mixed culture compared to the monoculture. For example, at 0.5 m it drops from 12.0 to 5.0 g dm^-3, and at 2.0 m it drops from 1.2 to 0.2 g dm^-3. Part c: Interspecific competition is shown because the root growth of L. tridentata is significantly restricted when grown with A. americana. Both species occupy overlapping niches and compete for limited soil resources (water, space, and mineral ions). The presence of the competitor limits the resources available, leading to reduced investment in or capacity for root biomass development in Larrea.
評分準則
Part a: [2 marks] - Award [1] for showing correct working: ((8.0 - 3.0) / 8.0) * 100. - Award [1] for correct final answer: 62.5% (accept 63%). Part b: [2 marks] - Award [1] for stating that the presence of Agave americana decreases the root density of Larrea tridentata at all distances. - Award [1] for supporting this with specific data from at least two points (e.g., decreased from 12.0 to 5.0 at 0.5 m and 8.0 to 3.0 at 1.0 m). Part c: [2 marks] - Award [1] for identifying that both species compete for the same limited soil resources (such as water or mineral nutrients). - Award [1] for explaining that competition leads to reduced resource availability, which suppresses the growth/root development of L. tridentata.
題目 2 · Data-based & Short-answer
6.25 分
An investigation was conducted on the metabolic rate of a desert rodent (Dipodomys merriami) across a range of ambient environmental temperatures. The oxygen consumption rate was measured as an indicator of metabolic rate.
Data Table: - Ambient Temperature (°C): 10, 20, 30, 40 - Oxygen consumption (cm^3 g^-1 h^-1): 4.5, 2.8, 1.2, 3.2
Questions: (a) Identify the ambient temperature that represents the most energy-efficient state (thermoneutral zone) for the rodent. [1] (b) Explain the physiological response and change in metabolic rate observed as the temperature decreases from 30°C to 10°C. [3] (c) Predict and explain the threat to the survival of the rodent if the ambient temperature is sustained at 42°C. [2]
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解題
Part a: The thermoneutral zone corresponds to the temperature with the lowest metabolic rate (lowest oxygen consumption), which is 30°C (1.2 cm^3 g^-1 h^-1). Part b: Dipodomys merriami is an endotherm and must maintain homeostasis (constant internal temperature). When the external temperature drops below its thermoneutral zone (from 30°C to 10°C), the rate of heat loss to the environment increases due to a steeper thermal gradient. To compensate and maintain core body temperature, the rodent increases physiological heat production (thermogenesis). This is achieved by increasing cell respiration rates (evidenced by the increase in oxygen consumption from 1.2 to 4.5 cm^3 g^-1 h^-1). Part c: If sustained at 42°C (above its core body temperature), the rodent is at risk of hyperthermia (overheating) and lethal dehydration. Since desert rodents have limited water resources, the active mechanisms used to dissipate heat (such as salivating or panting) consume precious water, and the excessive heat will cause cellular proteins and enzymes to denature, resulting in organ failure and death.
評分準則
Part a: [1 mark] - Award [1] for 30°C. Part b: [3 marks] - Award [1] for recognizing the rodent is an endotherm / maintains a constant body temperature. - Award [1] for stating that lower environmental temperatures increase the rate of heat loss (due to a steeper temperature gradient). - Award [1] for explaining that metabolic rate/cell respiration must increase to generate metabolic heat (to balance heat loss). Part c: [2 marks] - Award [1] for predicting hyperthermia / dehydration / enzyme denaturation / death. - Award [1] for explaining that temperatures above the core body temperature exceed the animal's cooling capacity and lead to rapid water loss or metabolic collapse.
題目 3 · Data-based & Short-answer
6.25 分
Yeast (Saccharomyces cerevisiae) cells were incubated in sealed flasks containing different carbon sources (Glucose, Galactose, or Lactose). The cumulative volume of carbon dioxide gas produced was measured over 30 minutes to compare respiration rates.
Data Table (CO2 produced in cm^3): - At 15 minutes: Glucose = 15.0; Galactose = 4.0; Lactose = 0.2 - At 30 minutes: Glucose = 35.0; Galactose = 12.0; Lactose = 0.5
Questions: (a) Calculate the rate of carbon dioxide production for yeast metabolizing glucose between 15 and 30 minutes. State the correct unit. [2] (b) Compare the suitability of glucose and galactose as substrates for cell respiration in yeast. [2] (c) Explain the lack of significant cellular respiration when yeast is incubated with lactose. [2]
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解題
Part a: Rate = (Change in CO2 volume) / (Change in time) = (35.0 - 15.0) / (30 - 15) = 20.0 cm^3 / 15 min = 1.33 cm^3 min^-1 (accept 1.3 or 1.33 cm^3 min^-1). Alternatively, if converted to hourly rate: 80 cm^3 h^-1. Part b: Glucose is a significantly more suitable and rapidly used substrate than galactose. At 15 minutes, glucose yields 15.0 cm^3 CO2 compared to only 4.0 cm^3 for galactose. By 30 minutes, glucose produces nearly three times as much CO2 (35.0 cm^3 vs 12.0 cm^3). Both can be respired, but glucose is metabolized much faster. Part c: Lactose is a disaccharide consisting of glucose and galactose. Yeast lacks the gene to produce lactase (beta-galactosidase), the enzyme needed to break the glycosidic bond in lactose. It also lacks specific transport proteins to move lactose into the cell. Thus, yeast cannot hydrolyze lactose into simple sugars to feed into the glycolysis pathway of cell respiration.
評分準則
Part a: [2 marks] - Award [1] for correct calculation: (35.0 - 15.0) / 15 = 1.33 (or 1.3). - Award [1] for correct units: cm^3 min^-1 (or cm^3/min). Accept 80 cm^3 h^-1. Part b: [2 marks] - Award [1] for stating that glucose is a more suitable substrate / produces a much higher rate of cell respiration than galactose. - Award [1] for supporting with comparative numbers (e.g., at 30 minutes, glucose produced 35.0 cm^3 of CO2 while galactose only produced 12.0 cm^3). Part c: [2 marks] - Award [1] for explaining that yeast lacks the enzyme lactase / beta-galactosidase to hydrolyze lactose. - Award [1] for explaining that without hydrolysis/transport, lactose cannot be broken down into monosaccharides (glucose/galactose) to enter glycolysis.
題目 4 · Data-based & Short-answer
6.25 分
The effect of a chemical inhibitor (Inhibitor X) on the rate of reaction of lactate dehydrogenase (LDH) was investigated across various substrate (lactate) concentrations.
Questions: (a) State the effect of increasing substrate concentration on the reaction rate in the presence of Inhibitor X. [1] (b) Deduce, with a reason based on the data, whether Inhibitor X is a competitive or non-competitive inhibitor. [3] (c) Describe how a non-competitive inhibitor affects both the maximum rate of reaction (Vmax) and the substrate concentration required to reach half Vmax (Km). [2]
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解題
Part a: As the substrate (lactate) concentration increases from 2 to 32 mmol dm^-3, the rate of reaction with Inhibitor X continuously increases from 5 to 47 micromol min^-1. Part b: Inhibitor X is a competitive inhibitor. In competitive inhibition, the inhibitor binds reversibly to the active site. At very high substrate concentrations, the substrate molecules outcompete the inhibitor molecules for the active site. The data shows that at 32 mmol dm^-3 substrate, the rate with the inhibitor (47 micromol min^-1) is nearly equal to the uninhibited rate (50 micromol min^-1). A non-competitive inhibitor would significantly lower the maximum rate (Vmax) regardless of substrate concentration. Part c: A non-competitive inhibitor binds to an allosteric site, altering the enzyme conformation. This decreases the maximum rate of reaction (Vmax) because some enzyme molecules are permanently incapacitated. It does not alter the affinity of the remaining active enzymes for the substrate, so the Michaelis constant (Km) remains unchanged.
評分準則
Part a: [1 mark] - Award [1] for stating that increasing substrate concentration increases the rate of reaction. Part b: [3 marks] - Award [1] for identifying it as a competitive inhibitor. - Award [1] for explaining that competitive inhibitors bind to the active site and can be outcompeted by high concentrations of substrate. - Award [1] for referencing data: at 32 mmol dm^-3, the inhibited rate (47 micromol min^-1) nearly reaches the maximum uninhibited rate (50 micromol min^-1). Part c: [2 marks] - Award [1] for stating that Vmax is decreased. - Award [1] for stating that Km remains unchanged (accept: affinity is unaffected).
卷二 甲部
Answer all compulsory structured questions in Section A.
5 題目 · 34 分
題目 1 · 結構題
6.8 分
An ecologist investigated the population size of a terrestrial snail species (Helix aspersa) in a local woodland using the mark-release-recapture method. On the first day, the ecologist captured, marked, and released 120 snails. One week later, 150 snails were captured, of which 30 were found to be marked. (a) Estimate the total population size of snails in the woodland using the Lincoln index formula. [2] (b) State two assumptions of the mark-release-recapture method that must be met for this estimate to be valid. [2] (c) Distinguish between density-dependent and density-independent factors that could limit this snail population, giving one example of each. [2.8]
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解題
(a) Lincoln index formula: \(N = \frac{n_1 \times n_2}{m_2}\) where \(n_1 = 120\), \(n_2 = 150\), and \(m_2 = 30\). Thus, \(N = \frac{120 \times 150}{30} = 600\). (b) Two assumptions of this method are: 1. There is no immigration or emigration of snails out of or into the woodland during the study period. 2. There are no births or deaths during the study period. (c) Density-dependent factors: Their impact increases as the population density increases (e.g., competition for food, disease, predation). Density-independent factors: Their impact is constant regardless of population density (e.g., extreme weather, floods, wildfires).
評分準則
(a) [2 marks] Award 1 mark for correct working (formula substitution) and 1 mark for correct calculation of 600. (b) [2 marks] Award 1 mark for each of two valid assumptions listed (max 2 marks). (c) [2.8 marks] Award 1 mark for defining density-dependent factors and 0.4 marks for a correct example. Award 1 mark for defining density-independent factors and 0.4 marks for a correct example.
題目 2 · 結構題
6.8 分
Thermoregulation is a vital homeostatic mechanism in humans to maintain a stable internal core body temperature. (a) Describe how the hypothalamus acts as the control center for thermoregulation. [2.8] (b) Explain the physiological responses of the body when core body temperature rises above the set point. [4]
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解題
(a) The hypothalamus contains thermoreceptors that monitor the temperature of the blood flowing through it. It also receives nerve impulses from peripheral thermoreceptors in the skin. It acts as a thermostat by comparing this sensory input to the biological set point of approximately 37 °C, coordinating corrective nervous and endocrine feedback mechanisms. (b) Responses to high core temperature: 1. Vasodilation: Arterioles leading to skin capillaries dilate, increasing blood flow to the skin surface to maximize heat loss via radiation. 2. Sweating: Sweat glands secrete sweat onto the skin surface, absorbing body heat to evaporate (evaporative cooling). 3. Flattening of hairs: Hair erector muscles relax, lowering body hairs to prevent trapping a layer of insulating air. 4. Metabolic adjustments: Metabolic rate may decrease to reduce internal heat generation.
評分準則
(a) [2.8 marks] Award 1 mark for stating that internal receptors in the hypothalamus monitor blood temperature, 1 mark for stating that it receives sensory input from skin thermoreceptors, and 0.8 marks for stating it coordinates the corrective responses to maintain the set point of ~37 °C. (b) [4 marks] Award 1 mark for each fully explained physiological response up to 4 marks: vasodilation (must mention arteriole dilation and increased blood flow to skin), sweating (must mention sweat secretion and evaporative cooling), flattening of hairs (must mention relaxation of erector muscles and reduction of insulating air), and metabolic/behavioral changes.
題目 3 · 結構題
6.8 分
Aerobic cell respiration involves a series of metabolic pathways occurring in different locations within the cell. (a) State the precise cellular location where each of the following stages occurs: Glycolysis, the Link reaction, and the Krebs cycle. [1.8] (b) Explain how the structure of the mitochondrion is adapted to its function in oxidative phosphorylation. [5]
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解題
(a) Glycolysis occurs in the cytoplasm (or cytosol). The Link reaction occurs in the mitochondrial matrix. The Krebs cycle occurs in the mitochondrial matrix. (b) Structural adaptations of the mitochondrion: 1. The inner mitochondrial membrane is folded into cristae to significantly increase the surface area available for the electron transport chain (ETC) proteins and ATP synthase. 2. The intermembrane space is extremely narrow, allowing for the rapid build-up of a high concentration of protons (\(H^+\)) to generate a steep electrochemical gradient. 3. The inner membrane contains specialized ATP synthase complexes that facilitate chemiosmosis. 4. The inner membrane is impermeable to protons, preventing passive leakage and maintaining the proton gradient. 5. The inner membrane contains the protein complexes of the electron transport chain to facilitate the transport of electrons and active pumping of protons.
評分準則
(a) [1.8 marks] Award 0.6 marks for each correct location: Glycolysis: Cytoplasm/Cytosol; Link reaction: Mitochondrial matrix; Krebs cycle: Mitochondrial matrix. (b) [5 marks] Award 1 mark for each correct adaptation up to 5: folding of inner membrane/cristae increases surface area, narrow intermembrane space accumulates protons rapidly/steep gradient, inner membrane contains ATP synthase, inner membrane is impermeable to protons, inner membrane contains ETC proteins/carriers.
題目 4 · 結構題
6.8 分
Enzymes are biological catalysts that speed up metabolic reactions. (a) Outline the differences between competitive and non-competitive enzyme inhibition. [3.8] (b) Explain how an increase in substrate concentration affects the rate of reaction in the presence of a competitive inhibitor versus a non-competitive inhibitor. [3]
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解題
(a) Differences: 1. Competitive inhibitors have a structure chemically similar to the substrate, whereas non-competitive inhibitors do not resemble the substrate. 2. Competitive inhibitors bind to the active site of the enzyme, while non-competitive inhibitors bind to an allosteric site (a site away from the active site). 3. Competitive inhibitors do not change the conformation of the active site, whereas non-competitive inhibitors change the conformation of the active site, preventing substrate binding. 4. Competitive inhibition can be overcome by high substrate concentrations, whereas non-competitive inhibition cannot. (b) Effects of substrate concentration: In the presence of a competitive inhibitor, increasing substrate concentration increases the reaction rate because the substrate outcompetes the inhibitor for the active site, allowing the reaction to reach the same maximum rate (\(V_{max}\)). In the presence of a non-competitive inhibitor, increasing substrate concentration does not overcome the inhibition because the inhibitor is not competing for the active site; thus, the maximum rate of reaction (\(V_{max}\)) is significantly reduced.
評分準則
(a) [3.8 marks] Award 1 mark for each of any three clear differences (max 3 marks), and 0.8 marks for pointing out the difference in active site conformational change. (b) [3 marks] Award 1.5 marks for explaining the effect on competitive inhibition (substrate outcompetes inhibitor, reaction can reach normal \(V_{max}\)), and 1.5 marks for explaining the effect on non-competitive inhibition (substrate does not compete, reaction rate remains lower and cannot reach normal \(V_{max}\)).
題目 5 · 結構題
6.8 分
The transmission of nerve impulses depends on the regulation of membrane potentials. (a) Describe how the resting membrane potential of an axon is maintained. [2.8] (b) Explain the sequence of events that occurs during depolarization and repolarization of an axon membrane during an action potential. [4]
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解題
(a) Resting potential maintenance: 1. The sodium-potassium pump (\(Na^+/K^+\) ATPase) actively pumps three sodium ions (\(Na^+\)) out of the axon for every two potassium ions (\(K^+\)) pumped in, using ATP. 2. The axon membrane is more permeable to potassium ions than sodium ions due to potassium leak channels, allowing potassium to diffuse out faster than sodium diffuses in. 3. Negatively charged organic ions (anions) remain trapped inside the cytoplasm of the axon, contributing to a resting membrane potential of approximately -70 mV. (b) Sequence of events: 1. A stimulus triggers the membrane potential to rise from resting potential to the threshold potential (approx. -55 mV). 2. Depolarization: Voltage-gated sodium (\(Na^+\)) channels open, causing a rapid influx of sodium ions into the axon down their concentration and electrochemical gradient, making the inside positive (+30 mV) relative to the outside. 3. Repolarization: Voltage-gated sodium channels close, and voltage-gated potassium (\(K^+\)) channels open. Potassium ions rapidly diffuse out of the axon down their gradient, restoring the negative internal membrane potential.
評分準則
(a) [2.8 marks] Award 1 mark for stating active transport of 3 \(Na^+\) out and 2 \(K^+\) in by the sodium-potassium pump. Award 1 mark for mentioning high membrane permeability to \(K^+\) due to leak channels. Award 0.8 marks for noting that trapped negative organic ions inside maintain the overall negative charge (~ -70 mV). (b) [4 marks] Award 1 mark for threshold potential trigger (-55 mV). Award 1.5 marks for describing depolarization (opening of voltage-gated sodium channels, rapid influx of \(Na^+\), internal positive charge). Award 1.5 marks for describing repolarization (closure of sodium channels, opening of voltage-gated potassium channels, efflux of \(K^+\), restoration of negative internal charge).
卷二 乙部
Answer one out of two extended-response questions. Quality of construction is worth 1 mark.
1 題目 · 16 分
題目 1 · extended-response
16 分
Section B: Extended Response
(a) Explain how blood glucose concentration is regulated in humans when it rises above normal levels. [4]
(b) Explain the role of the hypothalamus, pituitary gland, and ADH in osmoregulation when the body is dehydrated. [7]
(c) Distinguish between negative feedback and positive feedback, using one physiological example for each. [4]
An additional 1 mark is awarded for the quality of communication in this response. [1]
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解題
Part (a): Blood glucose regulation
When blood glucose levels rise above normal (for example, after a meal), beta cells in the pancreatic islets of Langerhans detect the increase and secrete insulin into the bloodstream. Insulin binds to specific receptors on target cells, particularly muscle cells and hepatocytes (liver cells). This binding triggers an increase in glucose uptake by these cells through the translocation of glucose transporter proteins to the cell membrane. Furthermore, insulin activates enzymes that convert glucose into glycogen for storage (a process called glycogenesis) and promotes the conversion of glucose into lipids. As glucose is removed from the blood and stored, blood glucose levels return to the normal homeostatic range.
Part (b): Osmoregulation during dehydration
Dehydration leads to an increase in blood solute concentration (hypertonicity) and a decrease in blood volume. This change is detected by specialized sensory receptors called osmoreceptors located in the hypothalamus. In response, the hypothalamus stimulates the posterior pituitary gland to release antidiuretic hormone (ADH), which was synthesized in the hypothalamus and stored in the pituitary. ADH is secreted into the bloodstream and travels to the kidneys, where it binds to receptors on the basolateral membrane of epithelial cells lining the collecting ducts. This binding initiates a second-messenger cascade that causes vesicles containing aquaporin water channels to fuse with the luminal (apical) membrane. The insertion of aquaporins significantly increases the water permeability of the collecting duct. As the filtrate passes down the collecting duct through the highly concentrated solute gradient of the renal medulla, water is reabsorbed by osmosis into the tissue fluid and back into the blood capillaries. Consequently, a small volume of highly concentrated urine is excreted, conserving water in the body.
Part (c): Negative vs. Positive Feedback
Negative feedback is a control mechanism where a change in a physiological variable triggers a response that counteracts or reverses the initial fluctuation, thereby maintaining a stable internal state (homeostasis). An example is thermoregulation, where an increase in body temperature triggers sweating and vasodilation to lower the temperature back to the set point. In contrast, positive feedback is a process where a change in a variable triggers a response that amplifies or reinforces the initial change, moving the system further away from its initial state until a definitive endpoint is reached. An example of positive feedback is the secretion of oxytocin during childbirth, where uterine contractions stimulate further oxytocin release, leading to even stronger contractions until delivery is complete.
Quality of Communication
The response is clearly structured into three distinct parts, employs accurate biological terms consistently (such as islets of Langerhans, glycogenesis, osmoreceptors, aquaporins, homeostasis), and demonstrates a logical progression of ideas from detection to effector response.
評分準則
Part (a): [Max 4 marks]
Beta cells in the islets of Langerhans (pancreas) detect high blood glucose levels.
Beta cells secrete insulin into the bloodstream.
Insulin binds to receptors on target cells (liver/muscle cells).
Stimulates increased uptake of glucose by target cells (facilitated diffusion via glucose transporters).
Stimulates glycogenesis / conversion of glucose to glycogen (in liver/muscle).
Stimulates conversion of glucose to fat/lipids.
Lowers blood glucose levels back to the normal set point (negative feedback).
Osmoreceptors in the hypothalamus detect the increase in osmolarity.
Hypothalamus stimulates the posterior pituitary gland.
Posterior pituitary gland releases antidiuretic hormone (ADH) into the blood.
ADH travels to the kidneys and binds to receptors on the collecting duct cells.
Triggers the fusion of vesicles containing aquaporins (water channels) with the cell membrane (luminal/apical membrane).
Increases the permeability of the collecting duct to water.
Water is reabsorbed from the filtrate into the hypertonic medulla / blood by osmosis.
Produces a small volume of concentrated urine.
Part (c): [Max 4 marks]
Negative feedback reverses/counteracts a change (to maintain a stable state/homeostasis), whereas positive feedback amplifies/reinforces a change (moving the system further from the initial state).
Accept one valid example of negative feedback (e.g., blood glucose regulation, thermoregulation, blood pH regulation, carbon dioxide levels).
Accept one valid example of positive feedback (e.g., oxytocin/uterine contractions during childbirth, blood clotting cascade, depolarization during action potential generation).
Award 1 mark for clear contrast in definition and up to 2 marks for explaining/describing the physiological examples correctly.
Quality of Construction (QC): [1 mark]
Award 1 mark for a well-structured, coherent response that uses appropriate biological terminology accurately and presents ideas in a logical sequence across all parts.
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