IB · BIOLOGY HL

Biology HL

Theme B: Form and function (HL) — Theme B HL

Name: ____________________Date: October 25, 2026

1.
Explain how the cohesion-tension theory accounts for the rise of water through the xylem of a tall tree.
[4 marks] · no calculator
Worked explanation
1. Evaporation from leaf mesophyll and diffusion out through stomata lower leaf water potential and create tension in the water column.
2. Hydrogen-bond cohesion transmits the pull down continuous xylem water columns, drawing water from roots rather than a pump lifting it at every point.
3. Adhesion to vessel walls helps support the column, while lignified walls resist collapse under negative pressure. The main energy input for transpiration comes from environmental heating, not xylem-cell ATP.
Common mistake: Cohesion-tension is a pulling mechanism; root pressure is not sufficient as the main explanation for water rising through a tall tree.
Marking points
- States that water evaporates from the mesophyll cells and leaves the stomata during transpiration, lowering water potential in the leaf.
- States that water molecules are cohesive, forming hydrogen bonds with each other, so a continuous unbroken column of water is pulled up the xylem.
- States that water molecules are adhesive to the walls of the narrow xylem vessels, helping counteract gravity.
- States that this creates tension (negative pressure) that is transmitted down the column, drawing water up from the roots.
Examiner tip: Cohesion, adhesion and tension must all be named explicitly — a vague description of 'water being pulled up' without these terms loses marks.
2.
Marking analysis: A learner attempts the following task: “Explain how the cohesion-tension theory accounts for the rise of water through the xylem of a tall tree.” Their response addresses only this point: “States that water evaporates from the mesophyll cells and leaves the stomata during transpiration, lowering water potential in the leaf.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that water evaporates from the mesophyll cells and leaves the stomata during transpiration, lowering water potential in the leaf.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that water molecules are cohesive, forming hydrogen bonds with each other, so a continuous unbroken column of water is pulled up the xylem.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that water molecules are adhesive to the walls of the narrow xylem vessels, helping counteract gravity.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this creates tension (negative pressure) that is transmitted down the column, drawing water up from the roots.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
3.
Outline the pressure flow (mass flow) hypothesis for translocation of sugars in phloem sieve tubes.
[4 marks] · no calculator
Worked explanation
1. At a source such as a photosynthesising leaf, sucrose loading lowers phloem water potential. Water enters from xylem by osmosis, raising hydrostatic pressure.
2. At a sink such as a growing root, sucrose unloading and use favour water exit and lower hydrostatic pressure.
3. The source-to-sink pressure gradient drives bulk sap flow through sieve tubes. Active loading uses ATP in species that employ it; the long-distance flow itself follows pressure, and some species load passively.
Common mistake: Phloem is not defined as downward transport: sources and sinks may be above or below each other.
Marking points
- States that sucrose enters sieve tubes at a source, lowering water potential. In an actively loading species, energy-dependent loading concentrates sucrose in the phloem; passive loading occurs in other species.
- States that water follows by osmosis from the xylem into the sieve tubes at the source, raising hydrostatic pressure.
- States that sucrose is unloaded at the sink (e.g. a growing root or fruit), and water follows it out by osmosis, lowering pressure there.
- States that the resulting pressure gradient from source to sink drives a mass flow of phloem sap along the sieve tubes.
Examiner tip: Always name a concrete source and sink; the hypothesis only makes sense in terms of a pressure gradient between two specific regions.
4.
Marking analysis: A learner attempts the following task: “Outline the pressure flow (mass flow) hypothesis for translocation of sugars in phloem sieve tubes.” Their response addresses only this point: “States that sucrose enters sieve tubes at a source, lowering water potential. In an actively loading species, energy-dependent loading concentrates sucrose in the phloem; passive loading occurs in other species.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that sucrose enters sieve tubes at a source, lowering water potential. In an actively loading species, energy-dependent loading concentrates sucrose in the phloem; passive loading occurs in other species.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that water follows by osmosis from the xylem into the sieve tubes at the source, raising hydrostatic pressure.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that sucrose is unloaded at the sink (e.g. a growing root or fruit), and water follows it out by osmosis, lowering pressure there.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the resulting pressure gradient from source to sink drives a mass flow of phloem sap along the sieve tubes.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
5.
A student measures the transpiration rate of a leafy shoot using a potometer at three different humidity levels. Predict and explain the effect of increasing humidity on the rate of water uptake.
[3 marks] · no calculator
Worked explanation
1. Higher external humidity reduces the water-vapour difference between the moist leaf air spaces and surrounding air, all else equal.
2. Less vapour diffuses out, reducing transpiration pull and usually the shoot's measured water uptake.
3. A potometer measures uptake, not evaporation directly: storage, growth and leaks can affect readings. Keep temperature, light, airflow and leaf area comparable, and do not invent numerical rates absent from the question.
Common mistake: Calling potometer uptake a direct transpiration measurement ignores other uses or storage of absorbed water.
Marking points
- Predicts that the rate of water uptake will decrease as humidity increases.
- Explains that a higher humidity reduces the water potential gradient between the air spaces inside the leaf and the surrounding air.
- Explains that this reduces the rate of diffusion of water vapour out of the stomata, slowing transpiration and hence water uptake.
Examiner tip: A potometer measures water uptake as a proxy for transpiration; always connect the two explicitly in your answer.
6.
Marking analysis: A learner attempts the following task: “A student measures the transpiration rate of a leafy shoot using a potometer at three different humidity levels. Predict and explain the effect of increasing humidity on the rate of water uptake.” Their response addresses only this point: “Predicts that the rate of water uptake will decrease as humidity increases.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: Predicts that the rate of water uptake will decrease as humidity increases.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that a higher humidity reduces the water potential gradient between the air spaces inside the leaf and the surrounding air.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that this reduces the rate of diffusion of water vapour out of the stomata, slowing transpiration and hence water uptake.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
7.
Explain why mature xylem water-conducting vessel elements are dead and hollow, whereas phloem sieve tube elements and their companion cells remain living in a flowering plant.
[4 marks] · no calculator
Worked explanation
1. Dead, hollow xylem vessel elements lose contents and end-wall obstructions, allowing low-resistance flow; lignified walls withstand transpiration tension.
2. Sieve tube elements remain living to maintain membranes and transport relationships, but have reduced organelles and no nucleus to leave room for sap.
3. Companion cells provide metabolic support through plasmodesmata and can support energy-dependent loading. Pressure drives long-distance sap flow; not every species uses active loading and not every xylem cell is dead.
Common mistake: Dead xylem vessels do not imply all cells in xylem tissue are dead; living parenchyma also occurs.
Marking points
- States that xylem transports water and dissolved minerals passively, driven by transpiration pull, which does not require living cell processes.
- States that the loss of cell contents and lignification of xylem cell walls creates open, reinforced tubes that offer minimal resistance to water flow and withstand tension without collapsing.
- States that living phloem cells maintain transport membranes and source-sink relationships; active loading, where it occurs, uses metabolic energy, whereas long-distance sap flow follows a pressure gradient.
- States that companion cells support sieve tube elements metabolically through plasmodesmata and, in actively loading species, supply the energy-dependent loading machinery.
Examiner tip: Compare conducting cell types, not every cell in either tissue: xylem also contains living parenchyma. Phloem bulk flow is pressure-driven, and loading can be active or passive depending on the species.
8.
Marking analysis: A learner attempts the following task: “Explain why mature xylem water-conducting vessel elements are dead and hollow, whereas phloem sieve tube elements and their companion cells remain living in a flowering plant.” Their response addresses only this point: “States that xylem transports water and dissolved minerals passively, driven by transpiration pull, which does not require living cell processes.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that xylem transports water and dissolved minerals passively, driven by transpiration pull, which does not require living cell processes.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the loss of cell contents and lignification of xylem cell walls creates open, reinforced tubes that offer minimal resistance to water flow and withstand tension without collapsing.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that living phloem cells maintain transport membranes and source-sink relationships; active loading, where it occurs, uses metabolic energy, whereas long-distance sap flow follows a pressure gradient.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that companion cells support sieve tube elements metabolically through plasmodesmata and, in actively loading species, supply the energy-dependent loading machinery.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
9.
A ring of bark and phloem is removed from around the trunk of a tree (girdling), leaving the xylem intact. Predict what will happen to the roots and explain your reasoning.
[3 marks] · no calculator
Worked explanation
1. Removing a continuous bark/phloem ring breaks the leaf-to-root sugar route. Roots below the ring lose a major carbon and respiratory-substrate supply.
2. Initially intact xylem can still carry water to leaves, so wilting need not be immediate and sugars may accumulate above the cut.
3. As root reserves run down, root growth and survival decline; later root failure can also compromise water supply. This prediction assumes no phloem bridge or alternative source restores continuity.
Common mistake: Immediate loss of all water transport is not the initial mechanism when the question explicitly leaves xylem intact.
Marking points
- Predicts that the roots will eventually die (or stop growing and be starved).
- Explains that girdling removes the phloem, which is the only tissue that transports sugars made by photosynthesis in the leaves down to the roots.
- Explains that the xylem remaining intact still supplies the leaves with water, so the leaves survive initially while the roots are cut off from their sugar supply.
Examiner tip: Girdling interrupts phloem continuity while initially sparing xylem transport. Phloem moves from sources to sinks, which may be above or below the source; xylem and phloem are not defined simply as upward versus downward transport.
10.
Marking analysis: A learner attempts the following task: “A ring of bark and phloem is removed from around the trunk of a tree (girdling), leaving the xylem intact. Predict what will happen to the roots and explain your reasoning.” Their response addresses only this point: “Predicts that the roots will eventually die (or stop growing and be starved).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: Predicts that the roots will eventually die (or stop growing and be starved).
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that girdling removes the phloem, which is the only tissue that transports sugars made by photosynthesis in the leaves down to the roots.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that the xylem remaining intact still supplies the leaves with water, so the leaves survive initially while the roots are cut off from their sugar supply.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
11.
Explain how the structure of guard cells enables stomata to open in response to light.
[4 marks] · no calculator
Worked explanation
1. Light activates guard-cell proton pumping; the resulting electrical conditions promote potassium uptake and solute accumulation.
2. Guard-cell water potential falls, so water enters by osmosis and turgor rises.
3. Uneven wall properties, including a thicker inner wall in the model described, make the paired cells bow apart rather than expand uniformly, opening the pore.
Common mistake: Ion accumulation alone does not mechanically open the pore; osmotic water entry and wall architecture connect the signal to movement.
Marking points
- States that light stimulates guard cells to actively pump H+ ions out, and K+ ions are then taken up into the guard cells.
- States that this lowers the water potential inside the guard cells, so water enters by osmosis and the guard cells become turgid.
- States that guard cell walls are unevenly thickened, being thicker on the inner side (adjacent to the pore) than the outer side.
- States that this uneven thickening causes the guard cells to bow outward as they become turgid, opening the stomatal pore between them.
Examiner tip: Both the ion-driven osmotic mechanism and the uneven-wall-thickness structural mechanism must be included for full marks; one alone is incomplete.
12.
Marking analysis: A learner attempts the following task: “Explain how the structure of guard cells enables stomata to open in response to light.” Their response addresses only this point: “States that light stimulates guard cells to actively pump H+ ions out, and K+ ions are then taken up into the guard cells.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that light stimulates guard cells to actively pump H+ ions out, and K+ ions are then taken up into the guard cells.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this lowers the water potential inside the guard cells, so water enters by osmosis and the guard cells become turgid.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that guard cell walls are unevenly thickened, being thicker on the inner side (adjacent to the pore) than the outer side.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this uneven thickening causes the guard cells to bow outward as they become turgid, opening the stomatal pore between them.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
13.
Outline one adaptation of xerophytic plants that reduces water loss, and explain how it achieves this.
[3 marks] · no calculator
Worked explanation
1. Sunken stomata form pits that shelter the pore from moving dry air and retain a more humid local atmosphere.
2. That reduces the leaf-to-air vapour gradient and slows diffusion of water vapour, lowering transpiration.
3. A thick cuticle or reduced leaf area is also valid if linked to its own permeability or area mechanism; one well-explained adaptation is sufficient.
Common mistake: A sunken stoma does not absorb extra water from dry air; it mainly reduces the gradient and exposure driving water loss.
Marking points
- Names a valid xerophytic adaptation, for example sunken stomata (in pits), a thick waxy cuticle, or reduced leaf surface area (e.g. spines instead of leaves).
- Explains a mechanism appropriate to the adaptation: sunken stomata trap humid air and reduce the water-vapour gradient, a thicker waxy cuticle reduces permeability to water, or reduced leaf area reduces the surface available for loss.
- Links the mechanism explicitly to a reduced rate of transpiration or evaporative water loss.
Examiner tip: Any one correctly explained adaptation earns full marks; do not list several superficially, as depth of explanation is credited over breadth.
14.
Marking analysis: A learner attempts the following task: “Outline one adaptation of xerophytic plants that reduces water loss, and explain how it achieves this.” Their response addresses only this point: “Names a valid xerophytic adaptation, for example sunken stomata (in pits), a thick waxy cuticle, or reduced leaf surface area (e.g. spines instead of leaves).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: Names a valid xerophytic adaptation, for example sunken stomata (in pits), a thick waxy cuticle, or reduced leaf surface area (e.g. spines instead of leaves).
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains a mechanism appropriate to the adaptation: sunken stomata trap humid air and reduce the water-vapour gradient, a thicker waxy cuticle reduces permeability to water, or reduced leaf area reduces the surface available for loss.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Links the mechanism explicitly to a reduced rate of transpiration or evaporative water loss.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
15.
Using the sliding filament model, outline the role of calcium ions in initiating muscle contraction.
[4 marks] · no calculator
Worked explanation
1. Excitation of the muscle fibre triggers calcium release from the sarcoplasmic reticulum into the sarcoplasm.
2. Calcium binds troponin, changing the troponin-tropomyosin complex so tropomyosin no longer blocks myosin-binding sites on actin.
3. Myosin heads can now form cross-bridges and undergo ATP-dependent cycling. Calcium exposes access; ATP hydrolysis supplies the energy for the cycle.
Common mistake: Calcium does not directly supply contraction energy or bind tropomyosin as the primary regulatory trigger; its binding target here is troponin.
Marking points
- States that a nerve impulse arriving at the neuromuscular junction triggers the release of calcium ions from the sarcoplasmic reticulum into the sarcoplasm.
- States that calcium ions bind to troponin on the actin filament.
- States that this binding causes a conformational change that moves tropomyosin away from the myosin-binding sites on actin.
- States that this exposes the binding sites, allowing myosin heads to attach to actin and form cross-bridges, initiating the power stroke that slides the filaments.
Examiner tip: Calcium's role is specifically to move tropomyosin off the actin binding sites via troponin — it does not directly power the contraction itself, which comes from ATP hydrolysis by myosin.
16.
Marking analysis: A learner attempts the following task: “Using the sliding filament model, outline the role of calcium ions in initiating muscle contraction.” Their response addresses only this point: “States that a nerve impulse arriving at the neuromuscular junction triggers the release of calcium ions from the sarcoplasmic reticulum into the sarcoplasm.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that a nerve impulse arriving at the neuromuscular junction triggers the release of calcium ions from the sarcoplasmic reticulum into the sarcoplasm.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that calcium ions bind to troponin on the actin filament.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this binding causes a conformational change that moves tropomyosin away from the myosin-binding sites on actin.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this exposes the binding sites, allowing myosin heads to attach to actin and form cross-bridges, initiating the power stroke that slides the filaments.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
17.
Explain why skeletal muscles function in antagonistic pairs, using the biceps and triceps at the elbow joint as an example.
[3 marks] · no calculator
Worked explanation
1. A contracting muscle pulls via tendons; it does not actively push a bone back when it relaxes.
2. Biceps contraction with triceps relaxation flexes the elbow. Triceps contraction with biceps relaxation extends it.
3. These opposite pulls control both directions around the joint, although real movement can also involve multiple cooperating muscles, gravity and elastic forces.
Common mistake: Relaxation is not an active pushing stroke; identify the opposing muscle that generates the returning pull.
Marking points
- States that muscles can only actively contract and pull; they cannot actively push or extend themselves.
- States that contraction of the biceps pulls the forearm up, flexing the elbow, while the triceps relaxes and is stretched.
- States that contraction of the triceps pulls the forearm down, extending the elbow, while the biceps relaxes and is stretched, so the two muscles working in opposition allow movement in both directions at the joint.
Examiner tip: At the elbow, opposing muscles provide flexion and extension because a contracting muscle pulls rather than pushes. Do not infer that every joint movement requires exactly one opposing pair: gravity, elastic recoil and multiple cooperating muscles can also contribute.
18.
Marking analysis: A learner attempts the following task: “Explain why skeletal muscles function in antagonistic pairs, using the biceps and triceps at the elbow joint as an example.” Their response addresses only this point: “States that muscles can only actively contract and pull; they cannot actively push or extend themselves.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that muscles can only actively contract and pull; they cannot actively push or extend themselves.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that contraction of the biceps pulls the forearm up, flexing the elbow, while the triceps relaxes and is stretched.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that contraction of the triceps pulls the forearm down, extending the elbow, while the biceps relaxes and is stretched, so the two muscles working in opposition allow movement in both directions at the joint.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
19.
Outline the process of ultrafiltration in the nephron, including the role of the pressure difference between the glomerulus and the Bowman's capsule.
[4 marks] · no calculator
Worked explanation
1. Resistance to blood leaving the glomerulus helps maintain high capillary hydrostatic pressure relative to Bowman's space; plasma oncotic pressure opposes filtration.
2. The net pressure drives water and small solutes across fenestrated endothelium, basement membrane and podocyte slits.
3. The barrier's size/charge selectivity retains cells and most large proteins. Glucose and urea can both enter filtrate, showing that filtration is not selective removal of only wastes.
Common mistake: Ultrafiltration is pressure-driven, not ATP-powered transport; useful glucose is filtered and subsequently reabsorbed.
Marking points
- States that the afferent arteriole entering the glomerulus is wider than the efferent arteriole leaving it, creating high hydrostatic pressure within the glomerular capillaries.
- States that net filtration pressure drives water and small solutes through the fenestrated endothelium, basement membrane and podocyte slit barrier into Bowman's space.
- States that the filtration barrier restricts cells and most plasma proteins according to properties including size and charge, retaining them in the blood.
- States that the fluid that enters the Bowman's capsule, called the filtrate, has essentially the same composition as blood plasma minus the large proteins and cells.
Examiner tip: Ultrafiltration is pressure-driven, not an active pump. Its barrier is selectively permeable by properties including size and charge; subsequent tubular reabsorption is selective and can be passive or energy-dependent.
20.
Marking analysis: A learner attempts the following task: “Outline the process of ultrafiltration in the nephron, including the role of the pressure difference between the glomerulus and the Bowman's capsule.” Their response addresses only this point: “States that the afferent arteriole entering the glomerulus is wider than the efferent arteriole leaving it, creating high hydrostatic pressure within the glomerular capillaries.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that the afferent arteriole entering the glomerulus is wider than the efferent arteriole leaving it, creating high hydrostatic pressure within the glomerular capillaries.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that net filtration pressure drives water and small solutes through the fenestrated endothelium, basement membrane and podocyte slit barrier into Bowman's space.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the filtration barrier restricts cells and most plasma proteins according to properties including size and charge, retaining them in the blood.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the fluid that enters the Bowman's capsule, called the filtrate, has essentially the same composition as blood plasma minus the large proteins and cells.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
21.
Explain how root pressure contributes to the movement of water into the xylem, and state one piece of evidence for its existence.
[3 marks] · no calculator
Syllabus placement evidence: B3.2.17 (AHL), printed p.60. Ion-driven positive root pressure is explicitly prescribed at HL. Guttation or cut-stem bleeding is an application as evidence, not a claim that root pressure drives most tall-tree transport.
Worked explanation
1. Living root tissues load ions into xylem, lowering its water potential and drawing in water osmotically.
2. With limited transpiration, incoming water can generate positive root pressure, pushing sap upward rather than pulling it under tension.
3. Guttation at leaf margins or bleeding from a cut stem supports positive pressure, especially when transpiration is low. It does not establish root pressure as the dominant mechanism in tall trees.
Common mistake: Guttation droplets are liquid xylem-derived water, not the water vapour lost in transpiration or condensed dew.
Marking points
- States that living root cells load mineral ions into the xylem, using active transport; the increased solute concentration lowers xylem water potential. Root hairs take ions up from the soil but do not directly line the xylem.
- States that water then moves into the xylem from the surrounding root cells by osmosis, generating a positive hydrostatic pressure that pushes water up the xylem from below.
- States valid evidence for root pressure, e.g. guttation (droplets of water forced out of leaf tips or edges at night when transpiration is low), or sap 'bleeding' from a cut stem.
Examiner tip: Root pressure is a secondary mechanism that supplements the cohesion-tension mechanism — it is most significant at night or under high humidity, when transpiration (and hence transpiration pull) is minimal.
22.
Marking analysis: A learner attempts the following task: “Explain how root pressure contributes to the movement of water into the xylem, and state one piece of evidence for its existence.” Their response addresses only this point: “States that living root cells load mineral ions into the xylem, using active transport; the increased solute concentration lowers xylem water potential. Root hairs take ions up from the soil but do not directly line the xylem.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that living root cells load mineral ions into the xylem, using active transport; the increased solute concentration lowers xylem water potential. Root hairs take ions up from the soil but do not directly line the xylem.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that water then moves into the xylem from the surrounding root cells by osmosis, generating a positive hydrostatic pressure that pushes water up the xylem from below.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States valid evidence for root pressure, e.g. guttation (droplets of water forced out of leaf tips or edges at night when transpiration is low), or sap 'bleeding' from a cut stem.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
23.
Outline the role of antidiuretic hormone (ADH) in regulating the water content of the blood.
[4 marks] · no calculator
Worked explanation
1. Concentrated blood has lower water potential, detected by hypothalamic osmoreceptors. ADH made in the hypothalamus is released from the posterior pituitary.
2. ADH increases aquaporin insertion in collecting-duct cells, raising water permeability. Water then moves down the existing osmotic gradient toward the medulla and blood.
3. Less water remains in urine, producing a smaller, more concentrated output and helping restore blood water balance. Reduced stimulus then decreases ADH release.
Common mistake: ADH does not actively pump water; it alters permeability so osmosis can recover more water.
Marking points
- States that a decrease in blood water potential (an increase in solute concentration) is detected by osmoreceptors in the hypothalamus.
- States that this stimulates the release of ADH from the posterior pituitary gland into the blood.
- States that ADH increases the permeability of the collecting duct walls to water, by inserting aquaporin channels.
- States that this increases water reabsorption back into the blood, producing a smaller volume of more concentrated urine and raising blood water potential back towards normal.
Examiner tip: ADH acts specifically on the collecting duct (and distal convoluted tubule), not the whole nephron — this is a classic example of negative feedback, since the hormone's effect (raising water potential) counteracts the original stimulus (falling water potential) that triggered its release.
24.
Marking analysis: A learner attempts the following task: “Outline the role of antidiuretic hormone (ADH) in regulating the water content of the blood.” Their response addresses only this point: “States that a decrease in blood water potential (an increase in solute concentration) is detected by osmoreceptors in the hypothalamus.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that a decrease in blood water potential (an increase in solute concentration) is detected by osmoreceptors in the hypothalamus.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this stimulates the release of ADH from the posterior pituitary gland into the blood.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that ADH increases the permeability of the collecting duct walls to water, by inserting aquaporin channels.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this increases water reabsorption back into the blood, producing a smaller volume of more concentrated urine and raising blood water potential back towards normal.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
25.
Outline how the loop of Henle creates a concentration gradient in the medulla that allows for the reabsorption of water from the collecting duct.
[5 marks] · no calculator
Worked explanation
1. The thick ascending limb reabsorbs NaCl by energy-dependent mechanisms while remaining relatively impermeable to water; the thin ascending segment, where present, moves salt passively.
2. The descending limb loses water to the increasingly concentrated medulla, concentrating fluid that then turns into the ascending limb.
3. Opposite flow and repeated local differences multiply the gradient along the loop; urea recycling also contributes, particularly deeper in the medulla.
4. The collecting duct passes through this gradient. When ADH raises permeability, water leaves by osmosis and the remaining urine becomes concentrated.
Common mistake: Water does not follow salt out of the ascending limb; its low water permeability is essential to the separation mechanism.
Marking points
- States that the water-impermeable thick ascending limb uses energy-dependent NaCl reabsorption to help establish the medullary gradient; the thin ascending limb, where present, transports salt passively.
- States that this creates an increasingly concentrated (hypertonic) environment in the medulla, with the highest solute concentration deep in the medulla.
- States that the descending limb is permeable to water, so water passively leaves it by osmosis into the concentrated medulla, further concentrating the fluid moving through the loop.
- States that this counter-current arrangement between the ascending and descending limbs multiplies and maintains a steep concentration gradient in the medulla tissue.
- States that as the collecting duct passes back through this concentrated medulla, water is drawn out of it by osmosis, especially when ADH increases its permeability, concentrating the urine.
Examiner tip: The countercurrent arrangement helps establish the gradient used by the collecting duct. Distinguish the thick and thin ascending limbs; urea recycling also contributes to medullary osmolality, especially in the inner medulla.
26.
Marking analysis: A learner attempts the following task: “Outline how the loop of Henle creates a concentration gradient in the medulla that allows for the reabsorption of water from the collecting duct.” Their response addresses only this point: “States that the water-impermeable thick ascending limb uses energy-dependent NaCl reabsorption to help establish the medullary gradient; the thin ascending limb, where present, transports salt passively.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that the water-impermeable thick ascending limb uses energy-dependent NaCl reabsorption to help establish the medullary gradient; the thin ascending limb, where present, transports salt passively.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this creates an increasingly concentrated (hypertonic) environment in the medulla, with the highest solute concentration deep in the medulla.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the descending limb is permeable to water, so water passively leaves it by osmosis into the concentrated medulla, further concentrating the fluid moving through the loop.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this counter-current arrangement between the ascending and descending limbs multiplies and maintains a steep concentration gradient in the medulla tissue.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that as the collecting duct passes back through this concentrated medulla, water is drawn out of it by osmosis, especially when ADH increases its permeability, concentrating the urine.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
27.
Describe the structure of a sarcomere, identifying the arrangement of actin and myosin filaments.
[4 marks] · no calculator
Worked explanation
1. A sarcomere runs from one Z line to the next. Thin actin filaments attach to those boundaries and extend inward.
2. Thick myosin filaments occupy the centre, overlapping actin toward either side. The A band spans the myosin length, the H zone contains myosin without actin overlap, and the I band contains thin filaments without myosin.
3. Contraction increases overlap and brings Z lines closer; filament lengths and the A-band length remain essentially constant while the H zone and I band narrow.
Common mistake: Actin and myosin slide; they do not shorten into smaller filaments during contraction.
Marking points
- States that a sarcomere is the repeating structural and functional unit of a myofibril, bounded at each end by a Z line.
- States that thin actin filaments are anchored to the Z lines and extend towards the centre of the sarcomere.
- States that thick myosin filaments are located in the centre of the sarcomere, overlapping with the actin filaments.
- States that the region containing only myosin (no overlapping actin) is the H zone, the whole myosin-containing region is the A band, and the region containing only actin is the I band.
Examiner tip: During contraction, the actin and myosin filaments themselves do not shorten — they simply slide further over one another, which is why the sarcomere as a whole shortens while the individual filament lengths stay constant.
28.
Marking analysis: A learner attempts the following task: “Describe the structure of a sarcomere, identifying the arrangement of actin and myosin filaments.” Their response addresses only this point: “States that a sarcomere is the repeating structural and functional unit of a myofibril, bounded at each end by a Z line.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that a sarcomere is the repeating structural and functional unit of a myofibril, bounded at each end by a Z line.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that thin actin filaments are anchored to the Z lines and extend towards the centre of the sarcomere.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that thick myosin filaments are located in the centre of the sarcomere, overlapping with the actin filaments.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the region containing only myosin (no overlapping actin) is the H zone, the whole myosin-containing region is the A band, and the region containing only actin is the I band.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
29.
Distinguish between slow oxidative (Type I) and fast glycolytic (Type IIx) muscle fibres, in terms of their structure and function. Do not generalise the glycolytic description to all Type II fibres: Type IIa fibres also have substantial oxidative capacity.
[4 marks] · no calculator
Course scope: Optional enrichment
Syllabus placement evidence: B3.3.1-B3.3.10, printed pp.60-61; B2.3.9, p.56. Motor units, sarcomeres and cardiac/striated adaptations are prescribed, but Type I-versus-Type IIx metabolic fibre classification is not. The required subtype comparison exceeds the listed outcomes.
Worked explanation
1. Type I fibres contain abundant mitochondria and myoglobin, supporting sustained aerobic ATP supply. Their slower contractions and fatigue resistance suit posture or endurance running.
2. Type IIx fibres have less oxidative machinery and depend strongly on rapid glycolytic ATP supply, supporting brief powerful contractions but faster fatigue.
3. This contrast is Type I versus IIx, not every Type II fibre: IIa fibres also have substantial oxidative capacity, and training can alter several fibre properties.
Common mistake: Treating all fast-twitch fibres as identical anaerobic cells ignores the oxidative capacity of Type IIa.
Marking points
- States that slow-twitch fibres contain more mitochondria and myoglobin, and rely mainly on aerobic respiration, contracting slowly but resisting fatigue for a long time.
- States that slow-twitch fibres are suited to endurance activities, e.g. long-distance running or maintaining posture.
- States that fast glycolytic Type IIx fibres have fewer mitochondria and less myoglobin than Type I fibres, rely strongly on anaerobic glycolysis, and contract quickly and powerfully but fatigue rapidly.
- States that fast glycolytic Type IIx fibres are suited to brief, high-power activity, such as sprinting or weightlifting.
Examiner tip: The proportion of slow-twitch to fast-twitch fibres in a muscle is largely genetically determined, which is part of why some athletes are naturally better suited to endurance sports and others to explosive power sports.
30.
Marking analysis: A learner attempts the following task: “Distinguish between slow oxidative (Type I) and fast glycolytic (Type IIx) muscle fibres, in terms of their structure and function. Do not generalise the glycolytic description to all Type II fibres: Type IIa fibres also have substantial oxidative capacity.” Their response addresses only this point: “States that slow-twitch fibres contain more mitochondria and myoglobin, and rely mainly on aerobic respiration, contracting slowly but resisting fatigue for a long time.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that slow-twitch fibres contain more mitochondria and myoglobin, and rely mainly on aerobic respiration, contracting slowly but resisting fatigue for a long time.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that slow-twitch fibres are suited to endurance activities, e.g. long-distance running or maintaining posture.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that fast glycolytic Type IIx fibres have fewer mitochondria and less myoglobin than Type I fibres, rely strongly on anaerobic glycolysis, and contract quickly and powerfully but fatigue rapidly.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that fast glycolytic Type IIx fibres are suited to brief, high-power activity, such as sprinting or weightlifting.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
31.
Explain how hydrostatic and oncotic pressure differences influence fluid filtration across systemic capillaries, and explain the role of the lymphatic system in returning filtered fluid to the blood.
[4 marks] · no calculator
Course scope: Optional enrichment
Syllabus placement evidence: B3.2.11-B3.2.13 (AHL), printed pp.59-60. Pressure filtration and lymph drainage are prescribed, but required oncotic-force balancing and the support explanation of glycocalyx/subglycocalyx exceed the guide account. The guide uses simplified venous reuptake; retain the qualified modern physiology as enrichment, not a newly prescribed model.
Worked explanation
1. Capillary hydrostatic pressure relative to interstitial pressure favours outward filtration, while the effective protein-related oncotic difference opposes it.
2. The endothelial glycocalyx and low-protein subglycocalyx space affect that oncotic gradient, so the classical plasma-versus-bulk-interstitial picture is incomplete.
3. Falling capillary pressure usually reduces filtration along most systemic vessels, rather than producing sustained universal venous-end reabsorption.
4. Lymphatics collect net filtered fluid and escaped proteins and return them to venous blood. Transient reabsorption and specialised absorbing beds are exceptions to the usual steady-state pattern.
Common mistake: A universal arterial-out/venous-in diagram cannot explain normal fluid return in most systemic tissues without lymphatics.
Marking points
- Explains that capillary hydrostatic pressure tends to drive fluid outward, opposed by interstitial hydrostatic pressure; the balance is a pressure difference across the barrier.
- Explains that plasma proteins generate an oncotic effect opposing filtration, with the endothelial glycocalyx and the low-protein space beneath it contributing to the effective oncotic gradient.
- States that declining capillary hydrostatic pressure generally reduces filtration along the vessel; sustained venous-end reabsorption is not the normal steady-state pattern in most systemic tissues.
- States that lymphatics collect net filtered fluid and escaped proteins and return them to the venous circulation.
Examiner tip: The classical arterial-filtration/venous-reabsorption diagram is a simplified historical model, not a universal steady-state description. Most systemic tissues rely on lymphatic return; transient reabsorption and specialised absorbing beds are exceptions.
32.
Marking analysis: A learner attempts the following task: “Explain how hydrostatic and oncotic pressure differences influence fluid filtration across systemic capillaries, and explain the role of the lymphatic system in returning filtered fluid to the blood.” Their response addresses only this point: “Explains that capillary hydrostatic pressure tends to drive fluid outward, opposed by interstitial hydrostatic pressure; the balance is a pressure difference across the barrier.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: Explains that capillary hydrostatic pressure tends to drive fluid outward, opposed by interstitial hydrostatic pressure; the balance is a pressure difference across the barrier.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that plasma proteins generate an oncotic effect opposing filtration, with the endothelial glycocalyx and the low-protein space beneath it contributing to the effective oncotic gradient.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that declining capillary hydrostatic pressure generally reduces filtration along the vessel; sustained venous-end reabsorption is not the normal steady-state pattern in most systemic tissues.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that lymphatics collect net filtered fluid and escaped proteins and return them to the venous circulation.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.
33.
Explain how changes in thoracic volume bring about inhalation, in terms of pressure changes described by Boyle's law.
[4 marks] · no calculator
Worked explanation
1. The diaphragm contracts downward and external intercostals lift the ribs, enlarging the thorax and expanding the lungs.
2. For a fixed gas amount at constant temperature, Boyle's law gives inverse pressure-volume dependence. Expansion transiently lowers alveolar pressure below atmospheric pressure.
3. Air flows down that pressure difference into the lungs until pressures equalise. The law is an explanatory approximation during an open airflow process; muscle contraction itself requires energy.
Common mistake: Airflow follows a pressure gradient, not a larger oxygen concentration gradient generated by the diaphragm.
Marking points
- States that contraction of the external intercostal muscles and the diaphragm increases the volume of the thoracic cavity.
- States Boyle's law: for a fixed amount of gas at constant temperature, pressure and volume are inversely related, so increasing the volume of the thoracic cavity decreases the air pressure inside the lungs.
- Explains that this creates a pressure inside the lungs that is lower than atmospheric pressure outside the body.
- Explains that air flows passively from the region of higher pressure (the atmosphere) to the region of lower pressure (the lungs), down this pressure gradient, resulting in inhalation.
Examiner tip: Airflow into the lungs is passive down an alveolar-to-atmospheric pressure gradient, but normal inhalation requires active respiratory-muscle contraction to expand the thorax. Do not describe the whole act of breathing as purely passive.
34.
Marking analysis: A learner attempts the following task: “Explain how changes in thoracic volume bring about inhalation, in terms of pressure changes described by Boyle's law.” Their response addresses only this point: “States that contraction of the external intercostal muscles and the diaphragm increases the volume of the thoracic cavity.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that contraction of the external intercostal muscles and the diaphragm increases the volume of the thoracic cavity.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: States Boyle's law: for a fixed amount of gas at constant temperature, pressure and volume are inversely related, so increasing the volume of the thoracic cavity decreases the air pressure inside the lungs.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that this creates a pressure inside the lungs that is lower than atmospheric pressure outside the body.
- Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that air flows passively from the region of higher pressure (the atmosphere) to the region of lower pressure (the lungs), down this pressure gradient, resulting in inhalation.
Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.