An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 HL (TZ1) IB Diploma Programme Biology paper. Not affiliated with or reproduced from IB.
Paper 1A
Answer all 40 multiple-choice questions. A calculator is required.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · multiple-choice
1 PastPaper.marks
Which of the following describes the physiological response to a rise in blood solute concentration (hyperosmolality) in humans?
A.Osmoreceptors in the hypothalamus shrink, leading to decreased ADH release from the posterior pituitary and less water reabsorption in the collecting ducts.
B.Osmoreceptors in the hypothalamus swell, leading to increased ADH release from the posterior pituitary and more water reabsorption in the collecting ducts.
C.Osmoreceptors in the hypothalamus shrink, leading to increased ADH release from the posterior pituitary and increased water reabsorption in the collecting duct.
D.Osmoreceptors in the hypothalamus swell, leading to decreased ADH release from the posterior pituitary and decreased water reabsorption in the collecting duct.
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PastPaper.workedSolution
Osmoreceptors in the hypothalamus detect a rise in blood solute concentration. As water leaves these cells by osmosis, they shrink, which fires nerve impulses triggering the release of ADH from the posterior pituitary. ADH travels to the kidneys, where it increases the water permeability of the collecting duct cells (via aquaporins), promoting water reabsorption into the blood.
PastPaper.markingScheme
Award [1] for the correct option C. Incorrect options: A (states decreased ADH release), B (states osmoreceptors swell and ADH increases), D (states osmoreceptors swell and ADH decreases).
PastPaper.question 2 · multiple-choice
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During the maintenance of the resting membrane potential in a neuron, what is the direct role of the sodium-potassium pump (\(\text{Na}^+/\text{K}^+\)-ATPase)?
A.It pumps three \(\text{Na}^+\) ions out of the cell and two \(\text{K}^+\) ions into the cell, using energy from ATP hydrolysis to maintain concentration gradients.
B.It allows three \(\text{Na}^+\) ions to diffuse into the cell and two \(\text{K}^+\) ions to diffuse out of the cell down their concentration gradients.
C.It pumps two \(\text{Na}^+\) ions out of the cell and three \(\text{K}^+\) ions into the cell to restore the membrane potential during hyperpolarization.
D.It opens in response to depolarization to allow rapid influx of \(\text{Na}^+\) ions into the axon.
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PastPaper.workedSolution
The active transport sodium-potassium pump moves 3 sodium (\(\text{Na}^+\)) ions out of the cell for every 2 potassium (\(\text{K}^+\)) ions moved into the cell, using energy from the hydrolysis of one ATP molecule. This maintains the high external concentration of \(\text{Na}^+\) and the high internal concentration of \(\text{K}^+\), which is critical for establishing the resting potential.
PastPaper.markingScheme
Award [1] for the correct option A. Reject B (passive diffusion), C (incorrect stoichiometry), D (describes voltage-gated sodium channels).
PastPaper.question 3 · multiple-choice
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A population of soil-dwelling beetles experiences a prolonged drought that reduces the abundance of soft-bodied prey. Only beetles with stronger mandibles can break open hard-shelled seeds to survive. Over several generations, the mean mandible strength of the population increases significantly. Which type of natural selection and evolutionary outcome does this represent?
A.Stabilizing selection, which maintains the intermediate phenotype and decreases genetic diversity.
B.Directional selection, where an extreme phenotype is favored and the allele frequency shifts over time.
C.Disruptive selection, where both extreme phenotypes are selected for, leading to speciation.
D.Frequency-dependent selection, where the rarest phenotype has the highest fitness.
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PastPaper.workedSolution
Directional selection occurs when environmental changes favor individuals at one extreme of the phenotypic range (in this case, stronger mandibles to crack seeds). This shifts the mean phenotype of the population towards that extreme over generations.
PastPaper.markingScheme
Award [1] for the correct option B. Reject A (stabilizing selection maintains the intermediate phenotype), C (disruptive selection favors both extremes, not just one), D (frequency-dependent is not shown here).
PastPaper.question 4 · multiple-choice
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During skeletal muscle contraction, what is the precise role of ATP hydrolysis?
A.It triggers the release of calcium ions from the sarcoplasmic reticulum.
B.It causes the tropomyosin-troponin complex to slide off the myosin-binding sites on the actin filament.
C.It provides the energy to cock the myosin head into its high-energy conformation, preparing it to form a cross-bridge.
D.It triggers the power stroke that slides the actin filament past the myosin filament.
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PastPaper.workedSolution
When ATP binds to the myosin head, it causes detachment of the myosin head from the actin filament. The hydrolysis of this ATP into ADP and inorganic phosphate (\(\text{P}_i\)) provides the energy to \"cock\" the myosin head back into its high-energy, ready position.
PastPaper.markingScheme
Award [1] for correct option C. Reject A (calcium release is mediated by action potential/voltage-gated channels in SR), B (calcium-troponin binding causes tropomyosin movement, not ATP hydrolysis), D (power stroke is triggered by the release of ADP and phosphate, not ATP hydrolysis).
PastPaper.question 5 · multiple-choice
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Which of the following correctly describes the structural organization of a eukaryotic nucleosome?
A.DNA wrapped twice around a core of eight histone proteins, stabilized by an H1 linker histone.
B.DNA wrapped once around a core of four histone proteins, stabilized by non-histone scaffold proteins.
C.RNA wrapped twice around a core of eight histone proteins, stabilized by ribosomal proteins.
D.DNA wrapped twice around a core of eight non-histone proteins, stabilized by an H1 linker histone.
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PastPaper.workedSolution
A nucleosome consists of DNA wound roughly twice (specifically ~1.67 times, or approximately 150 base pairs) around an octamer core of eight histone proteins (two each of H2A, H2B, H3, and H4). It is secured/stabilized by the H1 linker histone.
PastPaper.markingScheme
Award [1] for correct option A. Reject B (incorrect protein number and wrapping), C (contains RNA instead of DNA), D (states non-histone proteins form the octamer core).
PastPaper.question 6 · multiple-choice
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What is a fundamental difference between the lytic and lysogenic cycles of a bacteriophage?
A.In the lytic cycle, the viral genome is integrated as a prophage into the host chromosome, whereas in the lysogenic cycle, the viral DNA remains separate.
B.In the lytic cycle, host cell lysis is delayed as the virus replicates slowly, whereas in the lysogenic cycle, the host cell is ruptured immediately.
C.In the lytic cycle, the host cell's machinery is hijacked immediately to produce new virions and lyse the cell, whereas in the lysogenic cycle, the viral genome replicates passively along with the host chromosome without destroying the cell.
D.In the lytic cycle, only RNA genomes are replicated, whereas in the lysogenic cycle, only DNA genomes are replicated.
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PastPaper.workedSolution
In the lytic cycle, phage DNA immediately hijacks host resources to produce viral proteins and genomes, leading to cell lysis and release of new virions. In the lysogenic cycle, the phage DNA integrates into the bacterial chromosome (as a prophage) and is replicated passively during cell division, causing no immediate harm to the host cell.
PastPaper.markingScheme
Award [1] for correct option C. Reject A (definitions of prophage integration are swapped), B (cell lysis is immediate in lytic and delayed/absent in lysogenic), D (both cycles can occur with DNA bacteriophages like lambda phage).
PastPaper.question 7 · multiple-choice
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Which level of protein structure is primarily stabilized by hydrogen bonds between the atoms of the polypeptide backbone, rather than the R-groups (side chains)?
A.Primary structure
B.Secondary structure
C.Tertiary structure
D.Quaternary structure
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PastPaper.workedSolution
Secondary structure (alpha helices and beta-pleated sheets) is characterized by regular folding patterns stabilized exclusively by hydrogen bonds between the carbonyl oxygen (\(\text{C}=\text{O}\)) of one peptide bond and the amide hydrogen (\(\text{N}-\text{H}\)) of another in the polypeptide backbone. Tertiary and quaternary structures are stabilized by interactions among the amino acid R-groups (side chains).
PastPaper.markingScheme
Award [1] for correct option B. Reject A (stabilized by covalent peptide bonds), C and D (stabilized primarily by R-group interactions, e.g., disulfide bridges, ionic bonds, hydrophobic interactions, hydrogen bonds between R-groups).
PastPaper.question 8 · multiple-choice
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Why do greenhouse gases, such as carbon dioxide and methane, contribute to global warming while nitrogen and oxygen do not?
A.Greenhouse gases absorb incoming short-wave solar radiation, whereas nitrogen and oxygen reflect it back into space.
B.Greenhouse gases absorb and re-emit outgoing long-wave infrared radiation due to their molecular structure, whereas nitrogen and oxygen do not absorb this wavelength.
C.Greenhouse gases react chemically with ozone in the stratosphere to produce heat, whereas nitrogen and oxygen remain inert.
D.Greenhouse gases absorb ultraviolet (UV) radiation from the Sun and convert it directly into heat in the troposphere.
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PastPaper.workedSolution
Greenhouse gases have molecular structures (triatomic or larger, like \(\text{CO}_2\) and \(\text{CH}_4\)) that allow them to vibrate and absorb long-wave infrared radiation emitted from the Earth's surface. They then re-emit this radiation in all directions, trapping heat in the atmosphere. Diatomic molecules like \(\text{N}_2\) and \(\text{O}_2\) do not absorb infrared radiation.
PastPaper.markingScheme
Award [1] for correct option B. Reject A (incoming solar radiation is short-wave, which mostly passes through greenhouse gases), C (this is chemically inaccurate), D (UV is mainly absorbed by ozone in the stratosphere, not by greenhouse gases in the troposphere to drive global warming).
PastPaper.question 9 · multiple-choice
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A patient experiences severe dehydration. Which of the following describes the correct response of the osmoreceptors, the hormone released, and its target organ to restore homeostatic balance?
A.Osmoreceptors in the hypothalamus detect high solute concentration; ADH is released from the posterior pituitary; target cells in the collecting duct increase water reabsorption.
B.Osmoreceptors in the medulla detect low solute concentration; ADH is released from the anterior pituitary; target cells in the distal convoluted tubule decrease water reabsorption.
C.Osmoreceptors in the hypothalamus detect high solute concentration; aldosterone is released from the adrenal cortex; target cells in the loop of Henle increase sodium excretion.
D.Osmoreceptors in the pituitary detect low solute concentration; ADH is released from the hypothalamus; target cells in the collecting duct increase permeability to urea.
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PastPaper.workedSolution
Dehydration increases blood solute concentration (osmolarity). Osmoreceptors in the hypothalamus detect this change, leading to the release of antidiuretic hormone (ADH) from the posterior pituitary gland. ADH travels in the blood to the kidneys, where it acts on the cells of the collecting ducts, increasing their permeability to water to promote water reabsorption.
PastPaper.markingScheme
Award [1] for the correct answer (A). Reject other options because they describe incorrect locations of osmoreceptors, incorrect pituitary lobes, or incorrect physiological effects of ADH or aldosterone.
PastPaper.question 10 · multiple-choice
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During the repolarisation phase of an action potential in a myelinated axon, what is the state of the voltage-gated channels and the net movement of ions?
A.Voltage-gated \(Na^+\) channels are closed; voltage-gated \(K^+\) channels are open; \(K^+\) ions diffuse out of the axon.
B.Voltage-gated \(Na^+\) channels are open; voltage-gated \(K^+\) channels are closed; \(Na^+\) ions diffuse out of the axon.
C.Voltage-gated \(Na^+\) channels are closed; voltage-gated \(K^+\) channels are open; \(K^+\) ions are pumped into the axon.
D.Voltage-gated \(Na^+\) channels are open; voltage-gated \(K^+\) channels are open; \(Na^+\) ions are pumped out of the axon.
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PastPaper.workedSolution
During repolarisation, voltage-gated sodium (\(Na^+\)) channels close (become inactivated), and voltage-gated potassium (\(K^+\)) channels open. This allows potassium ions to rapidly diffuse out of the axon down their electrochemical gradient, restoring the negative resting membrane potential.
PastPaper.markingScheme
Award [1] for the correct answer (A). Reject B because sodium channels are closed and sodium does not diffuse out in this phase. Reject C and D because potassium ions diffuse down their concentration gradient, rather than being actively pumped during the repolarisation phase itself.
PastPaper.question 11 · multiple-choice
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A species of grass grows on soil contaminated with copper mining waste as well as on surrounding uncontaminated soil. Plants on the contaminated soil have evolved high tolerance to copper, while those on uncontaminated soil have low tolerance. Copper-tolerant plants grow more slowly on uncontaminated soil than non-tolerant plants. Which statement correctly explains this scenario?
A.Copper tolerance is a dominant trait that spreads through the population regardless of natural selection.
B.Copper-tolerant plants are outcompeted on uncontaminated soil because the metabolic cost of tolerance reduces their fitness where copper is absent.
C.The contaminated soil induced mutations in the plants to make them copper-tolerant.
D.Gene flow between the two populations will eventually lead to the extinction of the non-tolerant plants.
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PastPaper.workedSolution
This scenario demonstrates natural selection and trade-offs. On uncontaminated soil, there is no copper toxicity, so the metabolic costs associated with the copper tolerance mechanism decrease the overall fitness of tolerant plants compared to non-tolerant ones. Therefore, non-tolerant plants outcompete them in the clean environment.
PastPaper.markingScheme
Award [1] for the correct answer (B). Reject A because dominance alone does not determine allele frequency changes without selection. Reject C because environmental factors do not direct beneficial mutations (Lamarckian misconception). Reject D because gene flow would tend to homogenise the populations, but selection maintains the divergence.
PastPaper.question 12 · multiple-choice
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During skeletal muscle contraction, which of the following events occurs immediately after calcium ions bind to troponin?
A.Tropomyosin moves, exposing the binding sites on actin.
B.ATP binds to the myosin head, causing it to detach from actin.
C.The sarcoplasmic reticulum actively pumps calcium ions back into its lumen.
D.The H-zone and I-bands expand in width.
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PastPaper.workedSolution
When calcium ions bind to troponin, troponin undergoes a conformational change that pulls tropomyosin away from the myosin-binding sites on the actin filament, exposing them and allowing the myosin heads to bind.
PastPaper.markingScheme
Award [1] for the correct answer (A). Reject B because ATP binding causes detachment of the myosin head, not immediate post-calcium binding action. Reject C because cross-bridge formation can only happen after tropomyosin has moved to expose the sites. Reject D because the H-zone and I-bands shorten during contraction, not expand.
PastPaper.question 13 · multiple-choice
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An analysis of a double-stranded DNA molecule shows that 32% of its nucleotides contain thymine. What is the expected percentage of nucleotides containing guanine in this molecule?
A.18%
B.32%
C.36%
D.68%
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PastPaper.workedSolution
According to Chargaff's rules for double-stranded DNA, \([A] = [T]\) and \([G] = [C]\). Since \(T = 32\%\), \(A\) must also be \(32\%\), making \(A + T = 64\%\). The remaining nucleotides must be \(G + C = 100\% - 64\% = 36\%\). Since \([G] = [C]\), the percentage of guanine is \(36\% / 2 = 18\%\).
PastPaper.markingScheme
Award [1] for the correct answer (A). Correctly apply the complementary base pairing rules: \([A] = [T]\) and \([C] = [G]\) to calculate \(18\%\).
PastPaper.question 14 · multiple-choice
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Which of the following is a characteristic feature of the lysogenic cycle of a bacteriophage, but not the lytic cycle?
A.Transcription of phage genes by host RNA polymerase.
B.Integration of the viral genome into the host bacterial chromosome.
C.Lysis of the host cell wall to release newly assembled virions.
D.Synthesis of viral capsid proteins using host ribosomes.
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PastPaper.workedSolution
In the lysogenic cycle, the viral DNA integrates directly into the host bacterial chromosome, becoming a prophage. In the lytic cycle, the phage DNA remains separate and replication leads directly to host cell lysis.
PastPaper.markingScheme
Award [1] for the correct answer (B). Reject A and D because both cycles involve transcription of viral genes and translation using host ribosomes at some stage. Reject C because lysis only occurs in the lytic cycle.
PastPaper.question 15 · multiple-choice
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Which level of protein structure is primarily stabilized by hydrogen bonds between the carbonyl oxygen and the amide hydrogen of the polypeptide backbone, and which structure does it form?
A.Secondary structure; alpha-helices and beta-pleated sheets.
B.Tertiary structure; disulfide bridges and hydrophobic interactions.
C.Quaternary structure; interactions between multiple polypeptide subunits.
D.Primary structure; linear sequence of amino acids joined by peptide bonds.
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PastPaper.workedSolution
The secondary structure of a protein is stabilized by hydrogen bonds formed between the polar groups of the polypeptide backbone (the carbonyl oxygen \(C=O\) and the amide group \(N-H\)). This hydrogen bonding results in regular local structures such as alpha-helices and beta-pleated sheets.
PastPaper.markingScheme
Award [1] for the correct answer (A). Reject B because tertiary structure is stabilized by R-group interactions (disulfide bonds, ionic bonds, hydrophobic interactions). Reject C because quaternary structure involves interactions between separate polypeptide chains. Reject D because primary structure is stabilized by covalent peptide bonds.
PastPaper.question 16 · multiple-choice
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During intensive physical exercise, the body coordinates a homeostatic response to meet the increased demand for oxygen in skeletal muscles. Which of the following correctly describes this integrated response?
A.Increased blood carbon dioxide concentration is detected by chemoreceptors, which signal the medulla oblongata to increase ventilation rate and heart rate.
B.Decreased blood oxygen concentration is detected by thermoreceptors, which signal the cerebellum to trigger vasodilation of skin arterioles.
C.High lactic acid concentration is detected by baroreceptors, which stimulate the parasympathetic nervous system to increase blood pressure.
D.Low blood glucose levels stimulate the pituitary gland to secrete insulin, which increases cellular respiration in skeletal muscle.
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PastPaper.workedSolution
During exercise, active cellular respiration increases blood carbon dioxide levels. This is detected by chemoreceptors (in the medulla oblongata, carotid, and aortic bodies), which send impulses to the medulla oblongata. The medulla oblongata coordinates an autonomic response to increase both ventilation rate and heart rate, enhancing oxygen delivery.
PastPaper.markingScheme
Award [1] for the correct answer (A). Reject B because baroreceptors detect blood pressure changes, not oxygen; and ADH acts to conserve water rather than redistribute blood to muscles. Reject C because carbon dioxide levels rise during exercise, not fall; and the autonomic (sympathetic), not somatic, nervous system coordinates these internal adjustments. Reject D because exercise decreases blood pH (makes it more acidic) and sympathetic stimulation of skeletal muscle blood vessels typically involves vasodilatory pathways.
PastPaper.question 17 · multiple-choice
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Which of the following describes the cellular response in the collecting duct cells of the kidney when there is an increase in blood solute concentration?
A.Decreased secretion of ADH, leading to exocytosis of aquaporins into the luminal membrane.
B.Increased secretion of ADH, leading to endocytosis of aquaporins from the luminal membrane.
C.Increased secretion of ADH, leading to exocytosis of aquaporins into the luminal membrane.
D.Decreased secretion of ADH, leading to endocytosis of aquaporins from the basolateral membrane.
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PastPaper.workedSolution
An increase in blood solute concentration (high osmolarity) is detected by osmoreceptors in the hypothalamus, which stimulates the pituitary gland to secrete more ADH (antidiuretic hormone). ADH binds to receptors on the basolateral membrane of collecting duct cells. This initiates a signal transduction cascade that results in the exocytosis of vesicles containing aquaporins into the luminal (apical) membrane. This increases the water permeability of the membrane, allowing more water to be reabsorbed back into the bloodstream.
PastPaper.markingScheme
Award 1 mark for the correct option (C). Correct option identifies increased ADH secretion and the subsequent exocytosis of aquaporins into the luminal membrane.
PastPaper.question 18 · multiple-choice
1 PastPaper.marks
During an action potential in a mammalian neurone, what causes the rapid depolarization of the membrane?
A.The active transport of sodium ions out of the neurone.
B.The diffusion of potassium ions out of the neurone through voltage-gated channels.
C.The active transport of potassium ions into the neurone.
D.The diffusion of sodium ions into the neurone through voltage-gated channels.
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PastPaper.workedSolution
Depolarization is characterized by the membrane potential becoming more positive. This is achieved when voltage-gated sodium channels open in response to reaching the threshold potential, allowing sodium ions (\(Na^+\)) to diffuse rapidly down their electrochemical gradient into the neurone.
PastPaper.markingScheme
Award 1 mark for the correct option (D). Correct option identifies the passive movement (diffusion) of sodium ions into the cell through voltage-gated channels.
PastPaper.question 19 · multiple-choice
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Which process is essential for natural selection to result in evolutionary change within a population?
A.Direct alteration of an individual's genome by environmental stressors during its lifetime.
B.Variation in reproductive success among individuals with different heritable traits.
C.Equal survival rates among all phenotypes within the population.
D.The development of completely new alleles in response to an immediate survival need.
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PastPaper.workedSolution
For natural selection to occur, there must be heritable variation. Environmental selection pressures lead to differential survival and reproductive success (variation in reproductive success) among individuals. Those with advantageous traits pass these heritable traits to their offspring, changing allele frequencies over generations.
PastPaper.markingScheme
Award 1 mark for the correct option (B). Correct option identifies differential reproductive success based on heritable traits as the core mechanism.
PastPaper.question 20 · multiple-choice
1 PastPaper.marks
During skeletal muscle contraction, what is the role of calcium ions (\(Ca^{2+}\))?
A.They bind to tropomyosin, causing it to expose the myosin-binding sites on actin.
B.They bind to troponin, causing tropomyosin to move and expose myosin-binding sites on actin.
C.They bind to the myosin head, triggering the hydrolysis of ATP.
D.They block the active sites on actin to allow the muscle to relax.
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PastPaper.workedSolution
Calcium ions released from the sarcoplasmic reticulum bind to troponin. This binding causes a conformational shift in the troponin-tropomyosin complex, moving tropomyosin away from the myosin-binding sites on the actin filaments, allowing cross-bridges to form.
PastPaper.markingScheme
Award 1 mark for the correct option (B). Correct option accurately describes the specific binding partner (troponin) and the resulting action (movement of tropomyosin to expose actin sites).
PastPaper.question 21 · multiple-choice
1 PastPaper.marks
A sample of double-stranded DNA is analyzed and found to contain 34% cytosine. What is the expected percentage of adenine in this sample?
A.16%
B.34%
C.68%
D.32%
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PastPaper.workedSolution
In double-stranded DNA, base pairing rules dictate that cytosine (C) pairs with guanine (G), so \(\%C = \%G = 34\%\). Together, C and G make up \(34\% + 34\% = 68\%\) of the total bases. The remaining bases (adenine and thymine) must make up \(100\% - 68\% = 32\%\). Since adenine (A) pairs with thymine (T), \(\%A = \%T\), meaning adenine makes up half of the remainder: \(32\% / 2 = 16\%\).
PastPaper.markingScheme
Award 1 mark for the correct calculation and option (A). [Method: G = 34%, so G + C = 68%. Remaining is A + T = 32%. Therefore, A = 16%]
PastPaper.question 22 · multiple-choice
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Which of the following is a characteristic feature of the lysogenic cycle of a bacteriophage, but NOT the lytic cycle?
A.Integration of the viral genome into the host bacterial chromosome as a prophage.
B.Immediate destruction of the host cell's DNA.
C.Rapid assembly of new viral capsids inside the host cytoplasm.
D.Lysis of the host cell membrane to release progeny virions.
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PastPaper.workedSolution
In the lysogenic cycle, the bacteriophage inserts its DNA into the host bacterium's genome, where it remains dormant as a prophage and replicates during normal host cell division. In contrast, the lytic cycle bypasses this integration, leading directly to transcription, translation, viral assembly, and immediate host cell lysis.
PastPaper.markingScheme
Award 1 mark for the correct option (A). Correct option identifies the formation of a prophage as a unique feature of lysogeny.
PastPaper.question 23 · multiple-choice
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What type of interactions or bonds are primarily responsible for stabilizing the alpha-helix (\(\alpha\)-helix) secondary structure of a polypeptide?
A.Covalent disulfide bridges between cysteine residues.
B.Hydrogen bonds between the amine and carboxyl groups of the peptide backbone.
C.Ionic bonds between oppositely charged R-groups.
D.Hydrophobic interactions between non-polar side chains.
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PastPaper.workedSolution
Secondary structure (including both alpha-helices and beta-pleated sheets) is maintained by hydrogen bonds between the polar groups (carbonyl oxygen and amide hydrogen) of the main peptide backbone, not the variable R-groups.
PastPaper.markingScheme
Award 1 mark for the correct option (B). Correct option identifies hydrogen bonding within the polypeptide backbone as the stabilizing force of secondary structure.
PastPaper.question 24 · multiple-choice
1 PastPaper.marks
During intense physical exercise, how do the respiratory and cardiovascular systems coordinate to maintain homeostasis?
A.Decreased blood pH is detected by chemoreceptors, triggering an increase in ventilation rate.
B.Increased blood pH is detected by baroreceptors, triggering a decrease in heart rate.
C.Decreased blood carbon dioxide levels trigger the cardiac center to slow down blood flow.
D.Decreased oxygen levels in the lungs cause the bronchioles to constrict.
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PastPaper.workedSolution
During exercise, increased rate of cellular respiration produces more carbon dioxide (\(CO_2\)). This reacts with water in blood plasma to form carbonic acid, which dissociates into hydrogen ions and bicarbonate, decreasing blood pH. Chemoreceptors in the medulla oblongata, carotid bodies, and aortic bodies detect this decrease in pH and send signals to increase ventilation rate and depth to expel the excess \(CO_2\).
PastPaper.markingScheme
Award 1 mark for the correct option (A). Correct option correctly identifies the detection of decreased blood pH by chemoreceptors leading to an increased ventilation rate.
PastPaper.question 25 · multiple-choice
1 PastPaper.marks
What is the direct, cellular effect of an increase in blood solute concentration (high osmolarity) on the cells of the collecting duct in the kidney?
A.Decreased exocytosis of vesicles containing aquaporins into the basolateral membrane.
B.Increased insertion of aquaporins into the apical (luminal) membrane of collecting duct cells.
C.Active transport of water molecules from the filtrate into the hypertonic medulla.
D.Increased transcription of the gene encoding ADH (vasopressin) in the cells of the collecting duct.
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PastPaper.workedSolution
High blood solute concentration is detected by osmoreceptors in the hypothalamus, triggering the release of antidiuretic hormone (ADH) from the posterior pituitary. ADH travels in the blood and binds to receptors on the basolateral membrane of the collecting duct cells. This triggers a secondary messenger cascade that stimulates the exocytosis of vesicles containing aquaporins, resulting in their insertion into the apical (luminal) membrane. This increases the permeability of the collecting duct to water, facilitating water reabsorption down the osmotic gradient into the hypertonic medulla.
PastPaper.markingScheme
Award 1 mark for selecting correct option B. - Reject A: Exocytosis of aquaporins increases, and it occurs at the apical (luminal) membrane, not the basolateral membrane. - Reject C: Water moves passively via osmosis through aquaporins, not by active transport. - Reject D: ADH is synthesized in the hypothalamus and released by the pituitary, not transcribed or produced in the collecting duct cells.
PastPaper.question 26 · multiple-choice
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Which of the following describes a physiological response coordinated by the transfer of signals from the hypothalamus when core body temperature falls below the set point?
A.Vasodilation of arterioles feeding the skin capillaries to radiate heat.
B.Decreased thyroxine secretion to slow down metabolic rate and conserve energy.
C.Shivering, where rapid skeletal muscle contraction generates heat through cellular respiration.
D.Increased sweat gland activity to cool the blood via evaporation.
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PastPaper.workedSolution
When the core body temperature falls below the normal set point, the thermoregulatory center in the hypothalamus coordinates mechanisms to generate and conserve heat. Shivering is an involuntary, rapid contraction of skeletal muscles. This physical activity increases the rate of cellular respiration within muscle fibers, releasing heat as a metabolic byproduct to help raise body temperature. Vasodilation and sweating are heat-loss mechanisms, and a decrease in thyroxine would lower the basal metabolic rate, which is counterproductive when cold.
PastPaper.markingScheme
Award 1 mark for selecting correct option C. - Reject A: Vasoconstriction of arterioles occurs, not vasodilation, to minimize heat loss. - Reject B: Thyroxine secretion would increase, not decrease, to raise metabolic rate over time. - Reject D: Sweat gland activity decreases to minimize evaporative cooling.
PastPaper.question 27 · multiple-choice
1 PastPaper.marks
During the repolarization phase of an action potential in a mammalian myelinated axon, what is the state of the voltage-gated ion channels?
A.Sodium channels are open; potassium channels are closed.
B.Sodium channels are inactivated; potassium channels are open.
C.Sodium channels are closed; potassium channels are closed.
D.Sodium channels are open; potassium channels are open.
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PastPaper.workedSolution
Repolarization is characterized by the return of the membrane potential toward the resting negative value. During this phase, voltage-gated sodium channels rapidly close and become inactivated (via the inactivation gate), preventing any further influx of positive sodium ions. Simultaneously, voltage-gated potassium channels open, allowing potassium ions to diffuse rapidly out of the intracellular fluid down their electrochemical gradient, restoring the negative membrane potential.
PastPaper.markingScheme
Award 1 mark for selecting correct option B. - Reject A and D: Sodium channels are inactive (closed) during repolarization. - Reject C: Potassium channels must be open to allow repolarization to proceed.
PastPaper.question 28 · multiple-choice
1 PastPaper.marks
What is the precise role of calcium ions \(\text{Ca}^{2+}\) during transmission across a chemical cholinergic synapse?
A.They enter the postsynaptic neuron to trigger an excitatory postsynaptic potential (EPSP).
B.They bind to ligand-gated ion channels on the postsynaptic membrane to open sodium channels.
C.They trigger the fusion of acetylcholine-containing vesicles with the presynaptic membrane.
D.They activate acetylcholinesterase in the synaptic cleft to degrade neurotransmitters.
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PastPaper.workedSolution
When an action potential arrives at the presynaptic knob, it depolarizes the membrane and opens voltage-gated calcium channels. Calcium ions diffuse rapidly down their concentration gradient into the presynaptic cytoplasm. The increase in intracellular calcium concentration causes synaptic vesicles containing acetylcholine to move toward and fuse with the presynaptic membrane, releasing the neurotransmitter into the synaptic cleft via exocytosis.
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Award 1 mark for selecting correct option C. - Reject A: Calcium ions do not enter the postsynaptic neuron to trigger the EPSP; sodium influx causes the depolarization of the postsynaptic membrane. - Reject B: Acetylcholine binds to ligand-gated postsynaptic receptors, not calcium ions. - Reject D: Acetylcholinesterase is an enzyme present in the cleft; its activation is not dependent on presynaptic calcium influx.
PastPaper.question 29 · multiple-choice
1 PastPaper.marks
A population of bacteria is exposed to a sublethal concentration of an antibiotic over many generations. Which statement correctly describes the evolutionary process occurring in this population?
A.The antibiotic induces specific mutations in the bacterial DNA that directly confer resistance.
B.Natural selection acts on existing genetic variation, increasing the frequency of pre-existing resistance alleles.
C.The individual bacteria gradually acclimatize to the antibiotic, passing on their acquired physical resistance to offspring.
D.Genetic drift randomly eliminates the sensitive bacteria, leaving only the resistant ones.
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PastPaper.workedSolution
Natural selection acts on existing genetic variations. In any bacterial population, random mutations occur spontaneously, resulting in a small number of individuals that may possess resistance alleles. When exposed to the antibiotic (which acts as a selective pressure), those bacteria with the resistance allele have a survival advantage. They survive and reproduce, passing on the resistance genes to their offspring, which increases the frequency of the resistance allele in the gene pool over time. The antibiotic does not 'induce' mutations specifically designed to resist it, nor do bacteria consciously acclimatize and pass on acquired physical adaptations.
PastPaper.markingScheme
Award 1 mark for selecting correct option B. - Reject A: Mutations occur randomly, and are not directed or induced by the antibiotic to serve a specific purpose. - Reject C: Acquired characteristics / acclimatization are not coded in the genome and cannot be inherited (Lamarckian evolution). - Reject D: Genetic drift refers to random changes in allele frequency, whereas this scenario represents a directional selection pressure.
PastPaper.question 30 · multiple-choice
1 PastPaper.marks
During the contraction of a skeletal muscle sarcomere, which of the following structural changes is observed?
A.The A band shortens, while the I band and H zone remain constant in width.
B.The actin and myosin filaments themselves shorten in length.
C.The distance between adjacent Z lines decreases, and the H zone shortens.
D.The myosin heads detach from actin only when calcium ions bind directly to them.
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PastPaper.workedSolution
According to the sliding filament theory, during skeletal muscle contraction, actin (thin) filaments are pulled along myosin (thick) filaments toward the center of the sarcomere (M line). This movement pulls the adjacent Z lines closer together, shortening the overall length of the sarcomere. The H zone (the region containing only myosin) and the I band (the region containing only actin) shorten in width as the filaments overlap more. The A band, which represents the length of the thick myosin filaments, remains completely unchanged.
PastPaper.markingScheme
Award 1 mark for selecting correct option C. - Reject A: The A band remains constant in width, while the I band and H zone shorten. - Reject B: The individual filaments do not change their physical length; they only slide past one another. - Reject D: Myosin head detachment requires the binding of ATP, not calcium. Calcium binds to troponin on the actin filament.
PastPaper.question 31 · multiple-choice
1 PastPaper.marks
Which of the following statements correctly describes the structural organization of a nucleosome in eukaryotic chromatin?
A.A double-stranded DNA molecule wrapped around a core of eight histone proteins, secured by an H1 histone.
B.A single-stranded RNA molecule wrapped around a core of four acidic proteins.
C.An octamer of histone proteins enclosed by a lipid bilayer to protect the genetic material.
D.A loop of DNA wound around a single large ribosomal protein complex during translation.
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PastPaper.workedSolution
In eukaryotic chromatin, DNA is packaged to fit inside the nucleus. A nucleosome consists of a double-stranded DNA molecule (approximately 146 base pairs) wrapped around an octamer core of eight basic histone proteins (two copies each of H2A, H2B, H3, and H4). This structure is clamped and stabilized by a ninth histone protein, the linker histone H1, which helps package the nucleosomes into a more condensed 30-nm fiber.
PastPaper.markingScheme
Award 1 mark for selecting correct option A. - Reject B: It involves double-stranded DNA, not single-stranded RNA, and histones are basic (positively charged) proteins, not acidic. - Reject C: Histones are not enclosed by a lipid bilayer; nucleosomes are macromolecular protein-DNA complexes in the nucleoplasm. - Reject D: Nucleosomes are involved in DNA packaging, not translation, and do not involve ribosomal proteins.
PastPaper.question 32 · multiple-choice
1 PastPaper.marks
Which of the following events is unique to the lysogenic cycle of a temperate bacteriophage, distinguishing it from the lytic cycle?
A.Synthesis of viral structural proteins using host ribosomes.
B.Integration of the viral genome into the host bacterial chromosome as a prophage.
C.Lysis of the host cell wall to release newly assembled virions.
D.Injection of viral nucleic acids across the bacterial cell envelope.
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PastPaper.workedSolution
Temperate bacteriophages can undergo both lytic and lysogenic cycles. In the lysogenic cycle, the viral DNA integrates into the host bacterial chromosome, becoming a latent form known as a prophage. The prophage is replicated passively each time the host bacterium divides, without causing immediate damage to the host. In contrast, the lytic cycle involves the rapid replication of viral components, assembly of new virions, and active lysis of the host cell.
PastPaper.markingScheme
Award 1 mark for selecting correct option B. - Reject A and C: These are key events in the lytic cycle. - Reject D: Injection of nucleic acids occurs at the initiation of both the lytic and lysogenic cycles.
PastPaper.question 33 · multiple-choice
1 PastPaper.marks
Under which conditions would the secretion of ADH (antidiuretic hormone) increase, and what is its specific effect on the collecting duct of the kidney?
A.High blood solute concentration; increases the permeability of the collecting duct to water by inserting aquaporins.
B.Low blood solute concentration; decreases the permeability of the collecting duct to water by removing aquaporins.
C.High blood solute concentration; decreases the active transport of sodium ions out of the collecting duct.
D.Low blood solute concentration; increases the active transport of water into the vasa recta.
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PastPaper.workedSolution
When blood solute concentration (osmolarity) is high, osmoreceptors in the hypothalamus detect this change and stimulate the posterior pituitary gland to secrete ADH. ADH travels in the bloodstream to the kidneys, where it binds to receptors on the collecting duct cells. This triggers a signaling cascade that inserts aquaporin water channels into the apical membrane, increasing water reabsorption back into the blood.
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Award 1 mark for the correct answer (A).
PastPaper.question 34 · multiple-choice
1 PastPaper.marks
During the repolarization phase of an action potential in a neurone, what is the state of the voltage-gated sodium channels and the voltage-gated potassium channels?
A.Sodium channels are closed or inactivated; potassium channels are open.
B.Sodium channels are open; potassium channels are closed.
C.Sodium channels are open; potassium channels are open.
D.Sodium channels are closed; potassium channels are closed.
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PastPaper.workedSolution
During repolarization, the membrane potential must return to its resting negative value. To achieve this, the voltage-gated sodium channels close/inactivate (preventing further sodium influx) and the voltage-gated potassium channels open, allowing potassium ions to diffuse out of the neurone down their electrochemical gradient.
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PastPaper.question 35 · multiple-choice
1 PastPaper.marks
In a population of bacteria exposed to a constant low concentration of an antibiotic, a mutation arises that provides resistance to the antibiotic but reduces the growth rate of the bacteria in the absence of the antibiotic. Which statement best predicts the evolutionary outcome of this population if the antibiotic is permanently removed from the environment?
A.The frequency of the resistant allele will decrease over generations because of a selective disadvantage in the absence of the antibiotic.
B.The resistant bacteria will become dominant because mutated alleles are always selected for.
C.The mutation rate will increase to produce a new gene that compensates for the slower growth.
D.Disruptive selection will occur, splitting the population into two distinct species.
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PastPaper.workedSolution
Natural selection acts on the phenotype based on environmental pressures. In the absence of the antibiotic, the resistance gene provides no survival benefit but incurs a metabolic/growth cost (reduced growth rate). Therefore, non-resistant bacteria will reproduce faster and outcompete the resistant ones, decreasing the frequency of the resistance allele.
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Award 1 mark for the correct answer (A).
PastPaper.question 36 · multiple-choice
1 PastPaper.marks
During skeletal muscle contraction, what is the direct role of calcium ions (\(\text{Ca}^{2+}\)) and ATP?
A.\(\text{Ca}^{2+}\) binds to troponin, exposing myosin-binding sites on actin; ATP hydrolysis provides energy for the cocking of the myosin head.
B.\(\text{Ca}^{2+}\) binds to tropomyosin, causing it to detach from actin; ATP binds to actin to cause the power stroke.
C.\(\text{Ca}^{2+}\) binds to myosin, activating its ATPase activity; ATP is used to pump calcium back into the sarcoplasmic reticulum during contraction.
D.\(\text{Ca}^{2+}\) diffuses into the T-tubules to trigger depolarization; ATP binds to troponin to release myosin.
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PastPaper.workedSolution
Calcium ions bind to troponin, which induces a conformational change in tropomyosin, exposing the active sites on actin. ATP is required for the detachment of the myosin head from actin and its hydrolysis provides energy to cock the myosin head back into its high-energy state.
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Award 1 mark for the correct answer (A).
PastPaper.question 37 · multiple-choice
1 PastPaper.marks
Which statement correctly describes the structure of a eukaryotic nucleosome?
A.DNA wrapped around an octamer of eight histone proteins, secured by an H1 histone.
B.RNA wrapped around an octamer of eight histone proteins, secured by a linker protein.
C.A single histone protein wrapped around eight double-stranded DNA molecules.
D.DNA wrapped around a scaffold of non-histone acidic proteins inside the nucleolus.
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PastPaper.workedSolution
A nucleosome consists of approximately 146 base pairs of DNA wrapped around an octamer of core histones (two each of H2A, H2B, H3, and H4). The structure is stabilized and secured at the entry/exit site by another histone protein called H1.
PastPaper.markingScheme
Award 1 mark for the correct answer (A).
PastPaper.question 38 · multiple-choice
1 PastPaper.marks
What is a key difference between the lytic cycle and the lysogenic cycle of a bacteriophage?
A.In the lysogenic cycle, the viral genome integrates into the host bacterial chromosome as a prophage, whereas in the lytic cycle, the host DNA is degraded and new virions are assembled rapidly.
B.In the lytic cycle, the virus remains dormant inside the host cell for many generations, while in the lysogenic cycle, the host cell lysis occurs immediately.
C.The lytic cycle only occurs in RNA viruses, whereas the lysogenic cycle is unique to DNA viruses.
D.In the lysogenic cycle, host ribosomes translate viral proteins directly without replication of the viral genome.
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PastPaper.workedSolution
During the lysogenic cycle, the bacteriophage DNA integrates into the host cell's chromosome (becoming a prophage) and replicates passively as the bacteria divides. In contrast, the lytic cycle involves active viral replication, destruction of host DNA, assembly of new virions, and lysis of the host cell to release the virions.
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Award 1 mark for the correct answer (A).
PastPaper.question 39 · multiple-choice
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Which level of protein structure is stabilized primarily by hydrogen bonds between the atoms of the polypeptide backbone, and what structures does it form?
A.Secondary structure; forms alpha-helices and beta-pleated sheets.
B.Primary structure; forms the linear sequence of amino acids joined by peptide bonds.
C.Tertiary structure; forms complex three-dimensional globular shapes through disulfide bridges.
D.Quaternary structure; forms multi-subunit complexes through hydrophobic interactions.
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PastPaper.workedSolution
The secondary structure is established by hydrogen bonding between the amine (N-H) group of one amino acid and the carboxyl (C=O) group of another amino acid along the polypeptide backbone. This regular folding results in characteristic shapes, such as alpha-helices and beta-pleated sheets.
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Award 1 mark for the correct answer (A).
PastPaper.question 40 · multiple-choice
1 PastPaper.marks
How do greenhouse gases in the atmosphere contribute to global warming?
A.They absorb longer wavelength infrared radiation emitted from the Earth's surface and re-radiate it in all directions, including back toward Earth.
B.They absorb short wavelength ultraviolet radiation from the sun and reflect it back to space.
C.They react with ozone to destroy the ozone layer, allowing more high-energy cosmic rays to reach the surface.
D.They trap warm air currents physically in the troposphere, preventing convection.
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PastPaper.workedSolution
Solar radiation of shorter wavelengths (visible light, UV) passes through the atmosphere and warms the Earth's surface. The Earth then re-emits this energy as longer wavelength infrared radiation. Greenhouse gases absorb these infrared wavelengths and re-emit them in all directions, including back to Earth, trapping heat within the atmosphere.
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Award 1 mark for the correct answer (A).
Paper 1B
Answer all questions based on the provided experimental databases.
4 PastPaper.question · 35 PastPaper.marks
PastPaper.question 1 · structured-database
8.75 PastPaper.marks
An experiment was conducted to investigate the role of aquaporin-2 (AQP2) water channels in the collecting ducts of mammalian kidneys. Two groups of mice were studied: wild-type (WT) mice and knockout (KO) mice lacking the functional gene for AQP2. Both groups were subjected to 24 hours of water deprivation. The plasma antidiuretic hormone (ADH) concentration and urine osmolarity were measured before (0 hours) and after (24 hours) water deprivation.
(a) State the effect of 24-hour water deprivation on plasma ADH concentration in wild-type (WT) mice. [1 mark] (b) Compare and contrast the response of WT and KO mice to water deprivation in terms of urine osmolarity. [3 marks] (c) Deduce, with reference to the data, the physiological function of the AQP2 water channel and explain why KO mice exhibit exceptionally high ADH levels after 24 hours. [4.75 marks]
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PastPaper.workedSolution
(a) 24 hours of water deprivation causes a significant increase in plasma ADH concentration in wild-type (WT) mice, rising from \(1.5 \pm 0.3\ pg\ mL^{-1}\) to \(8.2 \pm 1.1\ pg\ mL^{-1}\).
(b) Compare and Contrast: - Similarities: Both WT and KO mice show an upward trend in urine osmolarity after 24 hours, although the change in KO is extremely small and within experimental error. - Differences: WT mice have a much higher starting urine osmolarity (\(1200\ mOsm\ L^{-1}\)) than KO mice (\(350\ mOsm\ L^{-1}\)). WT mice show a massive increase of \(1900\ mOsm\ L^{-1}\) (reaching \(3100\ mOsm\ L^{-1}\)), while KO mice show virtually no change (only an increase of \(10\ mOsm\ L^{-1}\) to \(360\ mOsm\ L^{-1}\)).
(c) Deduction and Explanation: - AQP2 channels are essential for water reabsorption from the kidney collecting ducts back into the bloodstream. - KO mice lack these channels, so they cannot reabsorb water, resulting in the production of highly dilute urine even under conditions of water deprivation (low urine osmolarity remains constant at ~360 mOsm/L). - This continuous loss of water leads to severe systemic dehydration and elevated blood solute concentration (hyperosmolarity) in KO mice. - High blood osmolarity stimulates osmoreceptors in the hypothalamus, triggering a continuous, compensatory, and highly elevated release of ADH (reaching \(24.5\ pg\ mL^{-1}\)) in an attempt to restore water balance, which is unsuccessful because the target aquaporins are absent.
PastPaper.markingScheme
(a) Award [1] for: - Clear statement of increase in plasma ADH concentration from 1.5 to 8.2 pg/mL (accept a general statement of a 5.5-fold increase).
(b) Award up to [3] (maximum [2] if only similarities or only differences are discussed): - WT starting/ending urine osmolarity is significantly higher than KO [1] - WT shows a major increase after water deprivation, whereas KO shows almost no significant change [1] - Both show a positive/increasing trend after 24 hours [1] - Correct supporting values from the table [1]
(c) Award up to [4.75]: - Deduces that AQP2 is responsible for water reabsorption from urine back into the blood [1] - Explains that without AQP2, water cannot be reabsorbed, leaving urine dilute [1] - Continuous water loss leads to dehydration/elevated blood solute concentration [1] - Dehydration acts as a persistent stimulus on the hypothalamus/pituitary gland, leading to hypersecretion of ADH in a failed compensatory feedback loop [1.75]
PastPaper.question 2 · structured-database
8.75 PastPaper.marks
Researchers isolated giant axons from the squid *Loligo pealeii* to investigate the mechanism of action of a newly discovered marine toxin, Toxin X. The isolated axons were bathed in physiological saline containing different concentrations of Toxin X. The resting membrane potential and the peak amplitude of the action potential were recorded following electrical stimulation.
(a) Describe the effect of Toxin X on the resting membrane potential and the action potential amplitude of the squid axon. [2 marks] (b) Suggest, with a reason, the specific target protein of Toxin X in the axonal membrane. [3.75 marks] (c) Explain the consequences of exposure to \(50\ \mu\text{M}\) Toxin X on synaptic transmission at the cholinergic neuromuscular junction. [3 marks]
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PastPaper.workedSolution
(a) Description: - Resting membrane potential remains constant / unaffected by Toxin X at approximately \(-70\ \text{mV}\) across all tested concentrations. - Action potential amplitude decreases progressively as the concentration of Toxin X increases, dropping from \(110\ \text{mV}\) in the control to \(0\ \text{mV}\) at \(50\ \mu\text{M}\).
(b) Specific target protein: - Voltage-gated sodium (\(\text{Na}^+\)) channels. Reasoning: - The generation of an action potential depends on the rapid influx of sodium ions through voltage-gated sodium channels during the depolarization phase. - A decreased amplitude indicates that fewer sodium channels are opening in response to depolarization. - At \(50\ \mu\text{M}\), the action potential is completely blocked (\(0\ \text{mV}\)), suggesting complete occlusion/inactivation of these channels. - The resting membrane potential is maintained by potassium leak channels and the sodium-potassium pump, which are clearly unaffected by the toxin since the resting potential remains constant.
(c) Consequences on synaptic transmission: - At \(50\ \mu\text{M}\), action potentials cannot propagate along the motor neuron axon. - When the electrical signal fails to reach the presynaptic terminal, depolarization-gated (voltage-gated) calcium channels do not open. - Without calcium influx, synaptic vesicles containing acetylcholine (ACh) cannot undergo exocytosis / fuse with the presynaptic membrane. - Consequently, no acetylcholine is released into the synaptic cleft, and the postsynaptic muscle fiber is not stimulated, causing flaccid paralysis.
PastPaper.markingScheme
(a) Award up to [2]: - Resting membrane potential is unaffected / remains stable at around \(-70\ \text{mV}\) [1] - Action potential amplitude decreases in a dose-dependent manner and is completely abolished at \(50\ \mu\text{M}\) [1]
(b) Award up to [3.75]: - Target identified as voltage-gated sodium channels [1] - Blocking sodium channels prevents the influx of sodium ions needed to depolarize the membrane during an action potential [1] - Resting potential channels (potassium leak channels/sodium-potassium pump) are unaffected, keeping resting potential stable [1.75]
(c) Award up to [3]: - Absence of action potential propagation to the axon terminal [1] - Failure to trigger voltage-gated calcium channel opening / calcium entry [1] - Prevention of acetylcholine release into the synaptic cleft, preventing muscle activation [1]
PastPaper.question 3 · structured-database
8.75 PastPaper.marks
An experiment was designed to observe the process of natural selection in real-time. Replicate populations of *Escherichia coli* were cultured in nutrient broth containing different concentrations of the antibiotic streptomycin over 50 generations. The population density (measured in colony-forming units per milliliter, \(\text{CFU mL}^{-1}\)) was determined at regular generation intervals.
(a) Identify the generation at which resistance to \(100\ \mu\text{g mL}^{-1}\) streptomycin first becomes detectably present in the population. [1 mark] (b) Explain the change in the population density of *E. coli* exposed to \(10\ \mu\text{g mL}^{-1}\) streptomycin from generation 0 to generation 50. [3.75 marks] (c) Describe how the principles of natural selection explain the observed patterns across all three experimental conditions. [4 marks]
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PastPaper.workedSolution
(a) Resistance to \(100\ \mu\text{g mL}^{-1}\) streptomycin first becomes detectably present in the population at Generation 20 (where the density rises from \(0.0\) to \(0.1 \times 10^7\ \text{CFU mL}^{-1}\)).
(b) Explanation: - At generation 0, the antibiotic streptomycin acts as a powerful selective agent, killing the vast majority of susceptible cells, leaving only a tiny density (\(4.2 \times 10^7\ \text{CFU mL}^{-1}\)). - Within the population, rare individuals already possessed random mutations conferring low-level resistance to streptomycin. - These resistant bacteria survived the antibiotic and reproduced via binary fission. - Over successive generations, the frequency of the resistant phenotype increased dramatically as susceptible strains were eliminated. - By generation 50, the population had fully adapted and reconstituted its maximum carrying capacity (\(85.1 \times 10^7\ \text{CFU mL}^{-1}\)), equivalent to the control group.
(c) Principles of Natural Selection: - Variation: Random mutations generate genetic variations (resistance/susceptibility) in the initial population. - Selection Pressure: Streptomycin acts as a selective filter. - At \(0\ \mu\text{g mL}^{-1}\) (Control), there is no selection pressure, so the population remains stable and high because no selective killing occurs. - At \(10\ \mu\text{g mL}^{-1}\), moderate selection pressure kills some cells, but low-level resistant mutants survive and propagate quickly, recovering by generation 30. - At \(100\ \mu\text{g mL}^{-1}\), severe selection pressure kills almost the entire population. Only extremely rare, high-level mutants survive. It takes more generations (until generation 20) for these individuals to multiply to detectable levels, but eventually, natural selection drives complete adaptation of the culture by generation 50.
PastPaper.markingScheme
(a) Award [1] for: - Generation 20.
(b) Award up to [3.75]: - Initial population drop because the antibiotic kills susceptible cells [1] - Variation exists where a few cells have pre-existing resistance mutations [1] - Surviving resistant cells reproduce/multiply over generations [1] - Resistance alleles increase in frequency in the gene pool, leading to population recovery [0.75]
(c) Award up to [4]: - Overproduction/competition exists, with antibiotic concentration acting as the primary selective pressure [1] - Control shows stable population because there is no selective advantage for mutant phenotypes [1] - Higher concentrations represent stronger selection, requiring more time/more specific mutations to adapt [1] - Explains the differential survival and adaptation of resistant strains over time across the concentrations [1]
PastPaper.question 4 · structured-database
8.75 PastPaper.marks
An in vitro biophysical study investigated the relationship between sarcomere length and isometric contractile tension in isolated vertebrate skeletal muscle fibers. The sarcomere length was set using a laser diffraction technique, and the maximum active tension generated upon electrical stimulation was measured.
(a) Identify the range of sarcomere lengths that yields maximum active tension. [1 mark] (b) Explain, in terms of the sliding filament model of muscle contraction, why active tension is \(0\%\) at a sarcomere length of \(3.6\ \mu\text{m}\). [3 marks] (c) Explain why the active tension decreases when the sarcomere length is reduced below \(2.0\ \mu\text{m}\). [2.75 marks] (d) State the role of ATP hydrolysis by myosin ATPase during the cross-bridge cycle. [2 marks]
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PastPaper.workedSolution
(a) The range of sarcomere lengths yielding maximum tension is \(2.0\ \mu\text{m}\) to \(2.2\ \mu\text{m}\).
(b) Explanation: - At \(3.6\ \mu\text{m}\), the sarcomere is highly stretched. - At this length, there is no physical overlap between the thin (actin) and thick (myosin) filaments. - Consequently, myosin heads cannot bind to actin binding sites to form cross-bridges. - Since no cross-bridges are formed, no sliding of filaments can occur, resulting in \(0\%\) tension.
(c) Explanation: - When sarcomere length is reduced below \(2.0\ \mu\text{m}\), the thin (actin) filaments from opposite ends of the sarcomere begin to overlap in the middle. - This overlapping interferes with the binding of myosin heads to actin, reducing effective cross-bridge formation. - Additionally, at extremely short lengths (such as \(1.5\ \mu\text{m}\)), the thick myosin filaments collide directly with the Z-discs, creating physical resistance/compression that opposes contraction.
(d) Role of ATP hydrolysis: - Binding of ATP causes detatchment of the myosin head from the actin filament. - Myosin ATPase hydrolyzes ATP into ADP and inorganic phosphate (\(\text{P}_i\)), releasing energy. - This energy causes the myosin head to change shape, "cocking" it into its high-energy conformation, which is required to bind to actin again for a subsequent power stroke.
PastPaper.markingScheme
(a) Award [1] for: - \(2.0\ \mu\text{m}\) to \(2.2\ \mu\text{m}\) (or any range specifying these boundaries).
(b) Award up to [3]: - Filaments are pulled entirely apart / lack of overlap between actin and myosin [1] - Myosin heads are unable to contact actin [1] - No cross-bridges can form / no mechanical force generated [1]
(c) Award up to [2.75]: - Overlap of actin filaments from opposite ends blocks/disrupts proper cross-bridge sites [1] - Myosin filaments collide physically with the Z-discs, creating opposing force [1] - Net decrease in the number of effective force-generating cross-bridges [0.75]
(d) Award up to [2]: - Provides energy to re-cock/reset the myosin head [1] - Puts the myosin head in the high-energy state necessary for binding to actin [1]
Paper 2 Section A
Answer all structured short-answer questions.
6 PastPaper.question · 48 PastPaper.marks
PastPaper.question 1 · structured-short-answer
8 PastPaper.marks
Osmoregulation is an essential homeostatic mechanism in mammals. (a) Outline the role of osmoreceptors in the hypothalamus when an individual is dehydrated. [2] (b) Explain how antidiuretic hormone (ADH) increases the permeability of the collecting duct to water. [4] (c) State two physiological or behavioural responses, other than ADH secretion, that are triggered to help restore water balance. [2]
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PastPaper.workedSolution
(a) Osmoreceptors in the hypothalamus detect a decrease in water potential (or an increase in the osmolarity/solute concentration) of the blood. Upon activation, they stimulate the posterior pituitary gland to release antidiuretic hormone (ADH) into the bloodstream and also stimulate the thirst center in the brain. (b) ADH travels via the blood to the kidneys, where it binds to specific receptors on the basolateral membrane of the collecting duct epithelial cells. This binding triggers an intracellular signaling cascade (involving G-proteins and cyclic AMP). This cascade causes vesicles containing water channel proteins called aquaporins to move to and fuse with the apical/luminal membrane of these cells. This insertion of aquaporins significantly increases the permeability of the membrane, allowing water to leave the collecting duct down its osmotic gradient into the hypertonic medulla, resulting in highly concentrated urine. (c) Other physiological and behavioural responses include: 1. Triggering of drinking behavior due to the sensation of thirst. 2. Vasoconstriction of blood vessels to help maintain blood pressure. 3. Decreased rate of sweating to conserve existing body water.
PastPaper.markingScheme
Part (a) [Max 2 marks]: - Osmoreceptors detect increased solute concentration/osmolarity (or decreased water potential) of blood. [1] - Stimulate the pituitary gland to release ADH. [1] - Trigger the sensation of thirst. [1]
Part (b) [Max 4 marks]: - ADH binds to specific receptors on collecting duct cell membranes. [1] - Triggers an intracellular/second messenger cascade (e.g., cAMP). [1] - Causes vesicles containing aquaporins to fuse with the luminal membrane. [1] - Increases the density of water channels/aquaporins in the membrane. [1] - Water is reabsorbed by osmosis out of the collecting duct (into the hypertonic medulla). [1]
Blood glucose concentration is regulated within tight physiological limits. (a) Compare the roles of alpha (\(\alpha\)) and beta (\(\beta\)) cells in the pancreatic islets. [2] (b) Explain the mechanism of action of insulin on target cells, such as liver and muscle cells. [4] (c) Distinguish between Type I and Type II diabetes in terms of their primary causes. [2]
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PastPaper.workedSolution
(a) Alpha cells detect low blood glucose concentrations and respond by secreting the hormone glucagon, which promotes glycogenolysis and gluconeogenesis to release glucose into the blood. Beta cells detect high blood glucose concentrations and respond by secreting insulin, which promotes cellular uptake of glucose and glycogen synthesis to reduce blood glucose levels. (b) When insulin is released, it binds to specific tyrosine kinase receptors on the cell surface membranes of target cells (such as hepatocyte liver cells and skeletal muscle fibers). This binding triggers a signal transduction cascade that stimulates intracellular vesicles carrying glucose transporter proteins (such as GLUT4) to migrate to and fuse with the plasma membrane. This significantly increases the rate of facilitated diffusion of glucose into the cells. Simultaneously, insulin activates intracellular enzymes that catalyze the conversion of glucose into glycogen (glycogenesis) for storage. (c) Type I diabetes is an autoimmune disease where the body's immune system mistakenly attacks and destroys the insulin-producing beta cells of the pancreas, leading to a complete lack of insulin production. Type II diabetes is characterized by insulin resistance, where target cells fail to respond effectively to insulin, often linked to lifestyle factors such as diet, obesity, and genetic predisposition.
Part (b) [Max 4 marks]: - Insulin binds to specific receptors on target cell (liver/muscle) membranes. [1] - Triggers a signal transduction pathway/chemical cascade. [1] - Causes GLUT4/glucose transporter vesicles to fuse with the plasma membrane. [1] - Increases the rate of facilitated diffusion of glucose into the cell. [1] - Activates enzymes that convert glucose to glycogen (glycogenesis). [1]
Part (c) [2 marks]: - Type I is caused by autoimmune destruction of beta cells (leading to an inability to produce insulin). [1] - Type II is caused by target cells failing to respond/becoming resistant to insulin. [1]
PastPaper.question 3 · structured-short-answer
8 PastPaper.marks
Neurons transmit electrical signals along their axons to communicate with other cells. (a) Explain how the resting membrane potential of \(-70\text{ mV}\) is established and maintained. [3] (b) Describe the membrane events that occur during the depolarization and repolarization phases of an action potential. [3] (c) State the role of calcium ions (\(\text{Ca}^{2+}\)) in synaptic transmission. [2]
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(a) The resting membrane potential of \(-70\text{ mV}\) is actively maintained by the sodium-potassium pump (\(\text{Na}^+/\text{K}^+\) ATPase), which uses ATP to actively pump three sodium ions out of the axon for every two potassium ions pumped in. This creates concentration gradients for both ions. Additionally, the resting membrane is highly permeable to potassium ions due to open potassium leak channels, allowing potassium to diffuse out of the cell down its gradient, while remaining highly impermeable to sodium. This net loss of positive charges, along with negatively charged proteins trapped inside, leaves the inside of the axon negatively charged relative to the outside. (b) When a stimulus depolarizes the membrane past a threshold value (typically \(-55\text{ mV}\)), voltage-gated sodium channels open rapidly. Sodium ions (\(\text{Na}^+\)) rush into the cell down their electrochemical gradient, causing the membrane potential to rise up to approximately \(+30\text{ mV}\). Following this, the voltage-gated sodium channels close (inactivate), and voltage-gated potassium channels open. Potassium ions (\(\text{K}^+\)) diffuse out of the axon down their gradient, removing positive charges from the cell and restoring the negative resting membrane potential. (c) When an action potential reaches the presynaptic terminal, it depolarizes the membrane and causes voltage-gated calcium channels to open. Calcium ions (\(\text{Ca}^{2+}\)) diffuse into the presynaptic bulb. This influx of calcium ions triggers synaptic vesicles containing neurotransmitters to migrate toward and fuse with the presynaptic membrane, releasing the neurotransmitter into the synaptic cleft via exocytosis.
PastPaper.markingScheme
Part (a) [Max 3 marks]: - Sodium-potassium pump actively transports three sodium (\(\text{Na}^+\)) ions out of the cell and two potassium (\(\text{K}^+\)) ions in. [1] - Creates concentration gradients for both ions. [1] - Potassium leak channels allow potassium to diffuse out of the axon, while sodium channels remain closed. [1] - Presence of negatively charged organic anions inside the axon contributes to the negative interior. [1]
Part (b) [Max 3 marks]: - Depolarization is caused by the opening of voltage-gated sodium channels, allowing sodium ions to rush into the axon. [1] - Repolarization occurs when sodium channels close and voltage-gated potassium channels open. [1] - Potassium ions diffuse out of the axon, restoring the negative membrane potential. [1]
Part (c) [2 marks]: - Action potential depolarization opens voltage-gated calcium channels, leading to an influx of calcium ions (\(\text{Ca}^{2+}\)) into the presynaptic neuron. [1] - Calcium ions stimulate synaptic vesicles to fuse with the presynaptic membrane and release neurotransmitters by exocytosis. [1]
PastPaper.question 4 · structured-short-answer
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Heavy metals such as copper and zinc are toxic to most plant species. However, some populations of Agrostis tenuis (bent grass) growing on soil contaminated by mining waste have developed tolerance to these metals. (a) State three conditions required for natural selection to occur in a population. [3] (b) Explain how the population of Agrostis tenuis evolved heavy metal tolerance over generations. [4] (c) Identify the type of selection (directional, stabilizing, or disruptive) that occurs when a population shifts towards a metal-tolerant phenotype. [1]
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(a) Natural selection requires: 1. Inherited variation within the population, meaning individuals possess different alleles for characteristics. 2. Overproduction of offspring leading to competition and a struggle for survival. 3. Environmental selection pressures that favor certain traits over others (differential survival and reproduction). (b) Within the ancestral population of Agrostis tenuis, random mutations created variation, with some individuals possessing alleles that conferred resistance/tolerance to heavy metals. When mining activities contaminated the soil, heavy metals acted as a powerful selective pressure. Non-tolerant plants suffered toxicity and died, failing to reproduce. In contrast, tolerant plants survived the toxic conditions, successfully reproduced, and passed their heritable alleles for metal tolerance on to their offspring. Over successive generations, the frequency of these advantageous metal-tolerance alleles increased significantly in the population's gene pool. (c) The evolution of metal tolerance represents directional selection, as the phenotypic character of the population shifted from one extreme (non-tolerant) toward the other extreme (highly tolerant) in response to a changing environment.
PastPaper.markingScheme
Part (a) [Max 3 marks]: - Inherited variation within the population. [1] - Overproduction of offspring leading to competition/struggle for survival. [1] - Selective pressure (differential survival/reproduction). [1] - Advantageous traits must be heritable (passed to offspring). [1]
Part (b) [Max 4 marks]: - Variation in heavy metal tolerance existed in the ancestral grass population (due to mutation). [1] - Contaminated soil acted as a strong selective agent/pressure. [1] - Non-tolerant plants died / failed to compete or reproduce. [1] - Tolerant individuals survived and successfully reproduced. [1] - Passed the advantageous alleles for metal tolerance to their offspring. [1] - Frequency of the metal-tolerance allele increased in the gene pool over generations. [1]
Part (c) [1 mark]: - Directional selection. [1]
PastPaper.question 5 · structured-short-answer
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Skeletal muscle contraction depends on the microscopic organization of protein filaments. (a) Outline the structural organization of a sarcomere in skeletal muscle. [2] (b) Explain the role of calcium ions (\(\text{Ca}^{2+}\)) and ATP in the sliding filament mechanism of muscle contraction. [4] (c) State the function of the sarcoplasmic reticulum during both muscle contraction and muscle relaxation. [2]
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(a) A sarcomere is the basic functional unit of a myofibril, defined as the region between two adjacent Z-lines. It consists of thin filaments composed mainly of the protein actin, which are anchored directly to the Z-lines, and thick filaments composed of the protein myosin, located in the central region of the sarcomere (the A-band). (b) Calcium ions (\(\text{Ca}^{2+}\)) bind to the protein troponin on the thin filament, causing a conformational change that moves tropomyosin away from the myosin-binding sites on the actin filament. ATP plays two essential roles: first, the binding of ATP to the myosin head causes it to detach from the actin filament, breaking the cross-bridge. Second, the hydrolysis of ATP to ADP and inorganic phosphate (Pi) provides the energy to 'cock' the myosin head into its high-energy state. The cocked head then binds to actin, and releasing ADP and Pi triggers the power stroke, pulling the thin filament toward the center of the sarcomere. (c) During muscle contraction, the sarcoplasmic reticulum releases stored calcium ions into the sarcoplasm via voltage-gated channels in response to an action potential propagating down the T-tubules. During muscle relaxation, the sarcoplasmic reticulum actively pumps calcium ions back into its lumen against their concentration gradient using ATP-dependent calcium pumps.
PastPaper.markingScheme
Part (a) [Max 2 marks]: - Bounded by Z-lines. [1] - Contains thin actin filaments (anchored to Z-lines) and thick myosin filaments (central). [1] - Shows alternating light/I-bands (actin only) and dark/A-bands (overlapping actin and myosin). [1]
Part (b) [Max 4 marks]: - Calcium ions bind to troponin. [1] - Causes tropomyosin to shift, exposing myosin-binding sites on actin. [1] - ATP binding causes the detachment of myosin heads from actin. [1] - ATP hydrolysis (to ADP + Pi) cocks the myosin head into a high-energy state. [1] - Power stroke occurs when myosin heads release ADP + Pi and pull actin inward. [1]
Part (c) [2 marks]: - Contraction: Releases calcium ions into the sarcoplasm. [1] - Relaxation: Actively pumps calcium ions back into its lumen. [1]
PastPaper.question 6 · structured-short-answer
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Viruses are unique, non-cellular agents of infection. (a) State two features of viruses that distinguish them from living organisms. [2] (b) Explain the roles of reverse transcriptase and integrase in the replication cycle of the Human Immunodeficiency Virus (HIV). [4] (c) Suggest why treating viral infections like HIV is more challenging than treating bacterial infections with antibiotics. [2]
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(a) Viruses are distinguished from living organisms because they: 1. Lack a cellular structure (they do not have cytoplasm, organelles, or a cell membrane). 2. Do not carry out metabolic processes or produce their own ATP. 3. Cannot reproduce independently, relying entirely on the host cell's machinery to replicate. (b) HIV is a retrovirus containing a single-stranded RNA genome. Reverse transcriptase is a viral enzyme that transcribes the single-stranded viral RNA into a complementary double-stranded DNA molecule inside the host cell's cytoplasm. Following reverse transcription, another viral enzyme called integrase binds to the viral DNA, transports it into the host cell's nucleus, and cuts the host chromosome to insert/integrate the viral DNA. Once integrated, the viral DNA is referred to as a provirus. (c) Treating viral infections is highly challenging because viruses replicate inside host cells, hijacking the host's own cellular machinery; therefore, drugs that disrupt viral replication often risk damaging host cells. Furthermore, retroviruses like HIV have extremely high mutation rates due to the lack of proofreading in reverse transcriptase, allowing them to rapidly evolve resistance to antiviral drugs.
PastPaper.markingScheme
Part (a) [Max 2 marks]: - Lack of cellular structure / non-cellular (no cytoplasm/organelles). [1] - Absence of independent metabolism / cannot generate ATP. [1] - Inability to reproduce outside of a host cell. [1] - Possess only one type of nucleic acid (either DNA or RNA, not both). [1]
Part (b) [4 marks]: - Reverse transcriptase converts viral RNA into double-stranded DNA. [1] - Occurs within the host cell cytoplasm. [1] - Integrase binds to the viral DNA and transports it into the host nucleus. [1] - Integrase inserts/integrates the viral DNA into the host genome/chromosome. [1]
Part (c) [Max 2 marks]: - Viruses live inside host cells and use host metabolic pathways, so targeting them can damage host tissues. [1] - Viruses do not have peptidoglycan cell walls or 70S ribosomes, rendering antibiotics ineffective. [1] - High mutation rate (especially RNA viruses) leads to rapid drug resistance. [1]
Paper 2 Section B
Answer any two out of the three extended-response questions. Quality of communication is assessed.
2 PastPaper.question · 32 PastPaper.marks
PastPaper.question 1 · extended-response
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a) Explain how the human body detects and responds to a decrease in core body temperature. [4]
b) Explain how antidiuretic hormone (ADH) controls water reabsorption in the kidney nephron. [5]
c) Discuss the role of negative feedback mechanisms in maintaining homeostasis, using the control of blood glucose concentration as an illustrative example. [6]
Up to 1 additional mark is available for the quality of communication. [1]
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a) Thermoregulation response: - Thermoreceptors in the hypothalamus (detecting blood temperature) and in the skin (detecting external temperature) monitor changes in core temperature. - When a decrease is detected, the hypothalamus acts as the control centre, sending nerve impulses to effectors. - Arterioles in the skin undergo vasoconstriction (shunting blood away from the superficial capillary beds to deep veins), which minimizes heat loss by radiation. - Shivering is triggered, involving rapid, involuntary skeletal muscle contractions that release heat energy as a byproduct of respiration. - Epinephrine/thyroxin may be released to increase basal metabolic rate and generate internal metabolic heat.
b) ADH action in the kidney: - Osmoreceptors in the hypothalamus monitor the osmolarity of the blood. - If blood solute concentration is high (dehydration), the posterior pituitary gland is stimulated to secrete ADH into the bloodstream. - ADH travels to the kidneys, where it binds to specific receptors on the basolateral membranes of cells lining the collecting ducts. - This binding activates a cascade of intracellular signals, causing vesicles containing aquaporins (water channel proteins) to fuse with the luminal (apical) membrane. - This dramatically increases the water permeability of the collecting duct cells. - Water moves out of the dilute filtrate in the collecting duct and into the hypertonic interstitial fluid of the renal medulla via osmosis, where it is reabsorbed into vasa recta capillaries.
c) Negative feedback and blood glucose: - Negative feedback is a regulatory process where a change in a physiological variable triggers a response that counteracts or reverses that change, returning the system to its optimal set point. - The normal blood glucose level is maintained around \(4.5\text{--}5.5\text{ mmol/L}\). - After eating, blood glucose levels rise; this is detected by beta cells in the pancreatic islets of Langerhans, which secrete insulin. - Insulin stimulates body cells to increase glucose uptake and triggers the liver/muscle cells to convert glucose into stored glycogen (glycogenesis), bringing blood glucose levels back down. - During fasting or strenuous exercise, blood glucose levels fall; this is detected by alpha cells in the islets of Langerhans, which secrete glucagon. - Glucagon stimulates the liver to break down stored glycogen into glucose (glycogenolysis) and release it into the blood, raising blood glucose levels back to the set point. - By continuously opposing fluctuations, this negative feedback system maintains a stable internal environment.
PastPaper.markingScheme
Part (a) [Max 4 marks]: - Thermoreceptors in hypothalamus/skin detect decrease in core temperature. (1 mark) - Hypothalamus coordinates nervous/hormonal signals to effectors. (1 mark) - Vasoconstriction of skin arterioles diverts blood flow away from skin surface, reducing heat loss. (1 mark; reject 'vasoconstriction of capillaries') - Shivering (involuntary muscle contraction) generates heat from cellular respiration. (1 mark) - Increase in metabolic rate/thyroxin release to increase heat production. (1 mark)
Part (b) [Max 5 marks]: - Osmoreceptors in the hypothalamus detect high blood osmolarity/solute concentration. (1 mark) - Posterior pituitary gland releases ADH into the blood. (1 mark) - ADH binds to receptors on the collecting duct (or distal convoluted tubule) cells. (1 mark) - Promotes the insertion of aquaporins (water channels) into the apical/luminal membranes. (1 mark) - Increases permeability of the collecting duct to water. (1 mark) - Water is reabsorbed by osmosis into the hypertonic medulla / blood, producing concentrated urine. (1 mark)
Part (c) [Max 6 marks]: - Definition: Negative feedback reverses a change to return a system to a set point. (1 mark) - High blood glucose (stimulus) is detected by beta cells in the pancreatic islets of Langerhans. (1 mark) - Beta cells secrete insulin. (1 mark) - Insulin stimulates glucose uptake by body cells / glycogenesis in liver and muscle cells, lowering blood glucose. (1 mark) - Low blood glucose (stimulus) is detected by alpha cells in pancreatic islets. (1 mark) - Alpha cells secrete glucagon. (1 mark) - Glucagon stimulates glycogenolysis (breakdown of glycogen to glucose) in the liver, raising blood glucose. (1 mark)
Quality of Communication [1 mark]: - Award 1 mark if the overall response is coherent, logically sequenced, and uses appropriate scientific vocabulary (e.g., vasoconstriction, islets of Langerhans, aquaporins, glycogenolysis) consistently across all three parts.
PastPaper.question 2 · extended-response
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a) Describe the structure of a skeletal muscle sarcomere, including how it appears under an electron microscope. [4]
b) Explain the role of calcium ions and ATP in the contraction of skeletal muscle. [6]
c) Distinguish between the structure and action of slow-twitch and fast-twitch muscle fibres. [5]
Up to 1 additional mark is available for the quality of communication. [1]
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a) Structure of a sarcomere: - A sarcomere is the repeating functional unit of a myofibril, located between two adjacent dark protein structures called Z-lines. - It contains two main types of protein filaments: thin filaments composed primarily of actin, and thick filaments composed primarily of myosin. - Actin filaments are anchored directly to the Z-lines and extend towards the centre of the sarcomere. - Under an electron microscope, alternating light and dark bands are visible: the light I-band contains only thin (actin) filaments, while the dark A-band represents the entire length of the thick (myosin) filaments. - In the middle of the A-band is the H-zone, which contains only thick (myosin) filaments and appears slightly lighter than the rest of the A-band where actin and myosin overlap. - The M-line is located in the exact centre of the H-zone, anchoring the thick filaments.
b) Role of calcium ions and ATP in contraction: - When an action potential depolarises the muscle cell membrane (sarcolemma), the electrical signal travels deep into the cell via T-tubules. - This depolarization triggers the sarcoplasmic reticulum to release stored calcium ions (\(\text{Ca}^{2+}\)) into the sarcoplasm. - Calcium ions bind to the protein troponin on the thin actin filaments. - This binding causes a conformational change in troponin, which pulls the regulatory protein tropomyosin off the myosin-binding sites on the actin filament, exposing them. - Myosin heads then bind to these exposed sites to form cross-bridges. - ATP is required for the cycle to continue: ATP binds to the myosin head, causing it to detach from the actin filament. - This ATP is then hydrolysed by the ATPase activity of the myosin head into ADP and inorganic phosphate (\(\text{P}_i\)), providing the energy to 'cock' the myosin head into its high-energy, primed state. - The cocked myosin head binds to a new position on the actin filament. Upon release of the phosphate and ADP, the myosin head undergoes a conformational change called the 'power stroke', pulling the actin filament toward the centre of the sarcomere and shortening the muscle.
c) Slow-twitch vs. fast-twitch fibres: - Myoglobin content: Slow-twitch fibres have high concentrations of myoglobin (giving them a red colour), whereas fast-twitch fibres have low myoglobin concentrations (appearing white). - Mitochondria: Slow-twitch fibres contain many mitochondria to support sustained aerobic respiration, whereas fast-twitch fibres contain fewer mitochondria and rely mainly on anaerobic glycolysis. - Capillary network: Slow-twitch fibres are surrounded by a dense network of capillaries to ensure a continuous supply of oxygen, whereas fast-twitch fibres have a much sparser capillary supply. - Contraction speed and force: Slow-twitch fibres contract slowly and produce less force but are highly resistant to fatigue, whereas fast-twitch fibres contract rapidly and powerfully but fatigue very quickly. - Glycogen stores: Fast-twitch fibres contain large stores of glycogen to fuel glycolysis, while slow-twitch fibres have lower glycogen stores, utilizing fatty acids and oxygen directly from the blood. - Function: Slow-twitch fibres are adapted for long-duration, endurance activities (such as maintaining posture or marathon running), whereas fast-twitch fibres are adapted for rapid, explosive movements (such as weightlifting or short sprints).
PastPaper.markingScheme
Part (a) [Max 4 marks]: - Sarcomere is defined as the functional unit of myofibril / distance between two Z-lines. (1 mark) - Actin forms the thin filaments; myosin forms the thick filaments. (1 mark) - I-band is the light band containing thin (actin) filaments only. (1 mark) - A-band is the dark band representing the entire span of thick (myosin) filaments. (1 mark) - H-zone is the lighter central region of the A-band containing only thick filaments. (1 mark)
Part (b) [Max 6 marks]: - Depolarization propagates down T-tubules, stimulating calcium release from the sarcoplasmic reticulum. (1 mark) - Calcium ions bind to troponin. (1 mark) - Troponin shifts, causing tropomyosin to move away from myosin-binding sites on actin. (1 mark) - Myosin heads bind to actin, forming cross-bridges. (1 mark) - ATP binds to myosin heads, causing them to detach from actin. (1 mark) - ATP is hydrolysed (to ADP and Pi) to cock/energize the myosin head. (1 mark) - Release of ADP and Pi triggers the power stroke, pulling actin towards the centre of the sarcomere. (1 mark)
Part (c) [Max 5 marks]: - Slow-twitch are red / have high myoglobin, whereas fast-twitch are white / have low myoglobin. (1 mark) - Slow-twitch have many mitochondria / rely on aerobic respiration, whereas fast-twitch have few mitochondria / rely on anaerobic glycolysis. (1 mark) - Slow-twitch have high capillary density, whereas fast-twitch have low capillary density. (1 mark) - Slow-twitch contract slowly and resist fatigue, whereas fast-twitch contract quickly and fatigue rapidly. (1 mark) - Slow-twitch have low glycogen stores, whereas fast-twitch have high glycogen stores. (1 mark) - Slow-twitch are used for endurance/posture, whereas fast-twitch are used for bursts of speed/strength. (1 mark)
Quality of Communication [1 mark]: - Award 1 mark if the candidate structures their answer clearly, distinguishing structural features from functional roles, and utilizes terms such as sarcomere, sarcoplasmic reticulum, troponin, and myoglobin correctly and consistently.