Biology SL

1. 1.

   A cubic cell has sides of length 2.0 cm. Calculate its surface-area-to-volume ratio. Compare it with a cubic cell of side 1.0 cm and explain one consequence for exchange with the environment.

   [5 marks] · no calculator

   Worked explanation

   1. A cube has six faces: for side 2.0 cm, area = 6(2.0 squared) = 24 cm^2 and volume = 2.0 cubed = 8.0 cm^3. The numerical SA:V is 24:8 = 3:1, or area/volume = 3.0 cm^-1.
   2. For side 1.0 cm, area = 6.0 cm^2 and volume = 1.0 cm^3, giving 6:1 or 6.0 cm^-1.
   3. The smaller cube has twice the exchange area per unit volume, so equivalent membrane flux can serve more cytoplasm relative to its demand; it does not have more total area.

   Common mistake: Comparing edge length directly with volume omits the six faces and does not calculate surface area.

   Marking points

   • For the 2.0 cm cube, calculates surface area = 24 cm².
   • Calculates volume = 8 cm³ and ratio = 3:1.
   • For the 1.0 cm cube, obtains ratio = 6:1.
   • States that the smaller cell has the greater surface-area-to-volume ratio.
   • Links the greater ratio to faster/more efficient exchange per unit volume.

   Examiner tip: A ratio must compare surface area and volume after both have been calculated.

2. 2.

   Marking analysis: A learner attempts the following task: “A cubic cell has sides of length 2.0 cm. Calculate its surface-area-to-volume ratio. Compare it with a cubic cell of side 1.0 cm and explain one consequence for exchange with the environment.” Their response addresses only this point: “For the 2.0 cm cube, calculates surface area = 24 cm².” 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: For the 2.0 cm cube, calculates surface area = 24 cm².
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates volume = 8 cm³ and ratio = 3:1.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: For the 1.0 cm cube, obtains ratio = 6:1.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the smaller cell has the greater surface-area-to-volume ratio.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Links the greater ratio to faster/more efficient exchange per unit volume.

   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. 3.

   Extension practice (includes HL content beyond the prescribed SL syllabus): Explain what happens to a plant cell placed in a hypotonic solution. Refer to water potential, osmosis and the cell wall.

   [4 marks] · no calculator

   Course scope: Optional enrichment

   Worked explanation

   1. The hypotonic solution has higher water potential than the cell contents, creating a tendency for net water entry through the partially permeable membrane.
   2. As water enters by osmosis, the vacuole expands and the protoplast presses against the cell wall.
   3. The resisting wall generates pressure that opposes further entry. The cell becomes turgid rather than lysing; water movement need not continue indefinitely.

   Common mistake: Saying the cell wall stops all water entry misses that entry occurs first and wall resistance then builds turgor.

   Marking points

   • The external solution has a higher water potential than the cell contents.
   • Water enters through the partially permeable membrane by osmosis.
   • The vacuole/cell contents expand and turgor pressure increases.
   • The rigid cell wall resists expansion, so the cell becomes turgid rather than bursting.

   Examiner tip: Extension practice: Osmosis and turgor are shared content (D2.3.2-D2.3.6), but the required explanation in water-potential terms is AHL. Guide references: D2.3.8-D2.3.11. This item is enrichment, not evidence of prescribed SL exam coverage. State the direction of water movement and the water-potential difference before describing the outcome.

4. 4.

   Marking analysis: A learner attempts the following task: “Extension practice (includes HL content beyond the prescribed SL syllabus): Explain what happens to a plant cell placed in a hypotonic solution. Refer to water potential, osmosis and the cell wall.” Their response addresses only this point: “The external solution has a higher water potential than the cell contents.” 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: The external solution has a higher water potential than the cell contents.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Water enters through the partially permeable membrane by osmosis.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: The vacuole/cell contents expand and turgor pressure increases.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: The rigid cell wall resists expansion, so the cell becomes turgid rather than bursting.

   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. 5.

   Extension practice (includes HL content beyond the prescribed SL syllabus): Outline the roles of the rough endoplasmic reticulum and the Golgi apparatus in the production and secretion of a protein.

   [4 marks] · no calculator

   Course scope: Optional enrichment

   Worked explanation

   1. A secreted protein is made by ribosomes associated with the rough ER and enters the ER lumen, keeping it within the secretory pathway rather than free in the cytosol.
   2. Transport vesicles carry it to the Golgi, where processing such as carbohydrate modification and sorting prepares it for its destination.
   3. A secretory vesicle buds from the Golgi and fuses with the plasma membrane. Its lumen opens to the exterior, releasing the protein by exocytosis without the protein crossing the lipid bilayer.

   Common mistake: Ribosomes synthesize protein; the Golgi processes and sorts it rather than translating mRNA.

   Marking points

   • States that ribosomes on the rough endoplasmic reticulum synthesize the protein, which enters the RER lumen.
   • States that the protein is transported (e.g. in vesicles) from the RER to the Golgi apparatus.
   • States that the Golgi apparatus modifies, processes and packages the protein (e.g. adding carbohydrate groups).
   • States that the Golgi apparatus packages the finished protein into a secretory vesicle, which moves to and fuses with the plasma membrane to release it by exocytosis.

   Examiner tip: Extension practice: Organelle recognition and general functions are shared content (A2.2.6, A2.2.11); the linked RER-Golgi protein-processing and secretion pathway is AHL. Guide references: B2.2.7-B2.2.9. This item is enrichment, not evidence of prescribed SL exam coverage. This pathway (ribosome → RER → Golgi → secretory vesicle → exocytosis) is a classic sequence question; learn the order of organelles involved.

6. 6.

   Marking analysis: A learner attempts the following task: “Extension practice (includes HL content beyond the prescribed SL syllabus): Outline the roles of the rough endoplasmic reticulum and the Golgi apparatus in the production and secretion of a protein.” Their response addresses only this point: “States that ribosomes on the rough endoplasmic reticulum synthesize the protein, which enters the RER lumen.” 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 ribosomes on the rough endoplasmic reticulum synthesize the protein, which enters the RER lumen.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the protein is transported (e.g. in vesicles) from the RER to the Golgi apparatus.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the Golgi apparatus modifies, processes and packages the protein (e.g. adding carbohydrate groups).
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the Golgi apparatus packages the finished protein into a secretory vesicle, which moves to and fuses with the plasma membrane to release it by exocytosis.

   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. 7.

   Explain how the fluid mosaic model describes the structure of the plasma membrane.

   [3 marks] · no calculator

   Worked explanation

   1. Phospholipids assemble with water-attracting heads facing the cytosol and extracellular fluid, and water-avoiding tails sheltered inside a bilayer.
   2. Different embedded and surface-associated proteins form the mosaic, supplying functions such as transport and receptors.
   3. Lipids and many proteins move laterally, making the membrane fluid rather than a rigid protein sandwich; anchoring restricts some proteins' mobility.

   Common mistake: Turning the hydrophobic tails toward the watery surroundings reverses the interactions stabilising the bilayer.

   Marking points

   • States that the membrane consists of a phospholipid bilayer, with hydrophilic heads facing outward and hydrophobic tails facing inward.
   • States that proteins are embedded within or attached to the bilayer, at varying depths and positions (a 'mosaic' of proteins).
   • States that phospholipids and many proteins can move laterally within the bilayer, giving it fluidity; some proteins have restricted mobility because they are anchored to other structures.

   Examiner tip: The name itself describes the model: 'fluid' refers to the constant movement of components, and 'mosaic' refers to the varied arrangement of embedded proteins.

8. 8.

   Marking analysis: A learner attempts the following task: “Explain how the fluid mosaic model describes the structure of the plasma membrane.” Their response addresses only this point: “States that the membrane consists of a phospholipid bilayer, with hydrophilic heads facing outward and hydrophobic tails facing inward.” 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 the membrane consists of a phospholipid bilayer, with hydrophilic heads facing outward and hydrophobic tails facing inward.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that proteins are embedded within or attached to the bilayer, at varying depths and positions (a 'mosaic' of proteins).
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that phospholipids and many proteins can move laterally within the bilayer, giving it fluidity; some proteins have restricted mobility because they are anchored to other structures.

   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. 9.

   Explain why a decreasing surface-area-to-volume ratio can limit the size of a cell that increases in size without changing its shape or exchange adaptations.

   [3 marks] · no calculator

   Worked explanation

   1. For an unchanged shape, multiplying cell length by two multiplies area by four but volume by eight. The exchange area per unit cytoplasm is therefore halved.
   2. Nutrient demand and waste production tend to scale with volume, whereas supply/removal across an unmodified membrane scales with area.
   3. Eventually exchange cannot meet demand quickly enough. Folding, shape change or internal transport can alter the constraint, so it is not a universal fixed size limit.

   Common mistake: A larger cell has more total surface area, not less; the shortage is surface area relative to volume.

   Marking points

   • States that as a cell increases in size, its volume (and hence its metabolic demand for nutrients and oxygen, and waste production) increases faster than its surface area.
   • States that exchange of materials with the environment occurs across the surface (plasma membrane), so a smaller surface-area-to-volume ratio limits the rate of exchange relative to the cell's needs.
   • States that beyond a certain size, the membrane surface area becomes insufficient to supply the cell's needs or remove its wastes fast enough, limiting maximum cell size.

   Examiner tip: For the same total volume and shape, many small cells provide more total surface area than one large cell. Cell shape, internal transport and membrane folding can modify this constraint; there is no single universal maximum cell size.

10. 10.

    Marking analysis: A learner attempts the following task: “Explain why a decreasing surface-area-to-volume ratio can limit the size of a cell that increases in size without changing its shape or exchange adaptations.” Their response addresses only this point: “States that as a cell increases in size, its volume (and hence its metabolic demand for nutrients and oxygen, and waste production) increases faster than its surface area.” 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 as a cell increases in size, its volume (and hence its metabolic demand for nutrients and oxygen, and waste production) increases faster than its surface area.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that exchange of materials with the environment occurs across the surface (plasma membrane), so a smaller surface-area-to-volume ratio limits the rate of exchange relative to the cell's needs.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that beyond a certain size, the membrane surface area becomes insufficient to supply the cell's needs or remove its wastes fast enough, limiting maximum cell size.

    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. 11.

    Outline primary active transport across a plasma membrane, and explain how ATP hydrolysis drives movement against a gradient.

    [3 marks] · no calculator

    Worked explanation

    1. A pump binds its transported substance on one side of the membrane. Moving it toward higher concentration, or against an ion's electrochemical gradient, requires energy.
    2. ATP hydrolysis is coupled to a change in the pump's conformation, exposing the binding site to the other side and releasing its cargo.
    3. Resetting the pump permits another cycle. This direct ATP coupling defines primary active transport, unlike passive diffusion or gradient-powered secondary transport.

    Common mistake: ATP does not make particles diffuse uphill; its hydrolysis must be coupled to the transport protein's cycle.

    Marking points

    • States that primary active transport uses specific membrane pumps to move a substance against its concentration gradient, or electrochemical gradient for ions.
    • States that movement against the relevant gradient requires an energy input.
    • States that ATP hydrolysis is coupled to changes in the pump's conformation that drive transport.

    Examiner tip: Primary active transport uses ATP directly. Secondary active transport uses an existing ion gradient; passive ion movement follows the electrochemical gradient, not necessarily the concentration gradient alone.

12. 12.

    Marking analysis: A learner attempts the following task: “Outline primary active transport across a plasma membrane, and explain how ATP hydrolysis drives movement against a gradient.” Their response addresses only this point: “States that primary active transport uses specific membrane pumps to move a substance against its concentration gradient, or electrochemical gradient for ions.” 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 primary active transport uses specific membrane pumps to move a substance against its concentration gradient, or electrochemical gradient for ions.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that movement against the relevant gradient requires an energy input.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that ATP hydrolysis is coupled to changes in the pump's conformation that drive transport.

    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. 13.

    Distinguish between a monosaccharide, a disaccharide and a polysaccharide, giving one example of each.

    [3 marks] · no calculator

    Worked explanation

    1. Glucose is one sugar unit, so it is a monosaccharide. Sucrose contains two joined sugar residues, so it is a disaccharide.
    2. Starch contains many glucose residues connected by glycosidic bonds, making it a polysaccharide. Hydrolysis can break those links to release smaller sugars.
    3. The distinction is the number of linked sugar units, not simply sweetness or whether a substance dissolves in water.

    Common mistake: Calling starch a chain of disaccharides misses that its repeating residues are monosaccharide units.

    Marking points

    • States that a monosaccharide is a single sugar unit, e.g. glucose.
    • States that a disaccharide is formed from two monosaccharides joined by a glycosidic bond (condensation reaction), e.g. sucrose.
    • States that a polysaccharide is a long chain of many monosaccharides joined together, e.g. starch or glycogen.

    Examiner tip: Glycosidic bonds join sugar residues and can be broken by hydrolysis using water. Condensation is a useful school-level model for bond formation, but actual cellular polysaccharide synthesis commonly transfers sugars from activated donors rather than directly releasing water from free monosaccharides.

14. 14.

    Marking analysis: A learner attempts the following task: “Distinguish between a monosaccharide, a disaccharide and a polysaccharide, giving one example of each.” Their response addresses only this point: “States that a monosaccharide is a single sugar unit, e.g. glucose.” 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 a monosaccharide is a single sugar unit, e.g. glucose.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a disaccharide is formed from two monosaccharides joined by a glycosidic bond (condensation reaction), e.g. sucrose.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a polysaccharide is a long chain of many monosaccharides joined together, e.g. starch or glycogen.

    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. 15.

    Extension practice (HL content beyond the prescribed SL syllabus): Outline the level of protein structure at which hydrogen bonding creates an alpha helix or beta pleated sheet.

    [2 marks] · no calculator

    Course scope: Optional enrichment

    Worked explanation

    1. An alpha helix or beta sheet is a regular local fold of the polypeptide backbone, so it is secondary protein structure.
    2. Hydrogen bonds between backbone C=O and N-H groups stabilise the repeated pattern; they are not the peptide bonds that establish the amino-acid sequence.

    Common mistake: Attributing these repeating hydrogen bonds only to R groups confuses secondary backbone folding with tertiary side-chain interactions.

    Marking points

    • States that this is the secondary structure of a protein.
    • States that hydrogen bonds form between the carbonyl (C=O) and amine (N–H) groups of the polypeptide backbone, creating a regular, repeating shape.

    Examiner tip: Extension practice: Hydrogen-bonded alpha helices and beta sheets as secondary protein structure are AHL. Guide references: B1.2.8. This item is enrichment, not evidence of prescribed SL exam coverage. Primary structure is the amino acid sequence, secondary is local hydrogen-bonded folding (helix/sheet), tertiary is the overall 3D shape from all bond types, and quaternary is multiple polypeptide chains together.

16. 16.

    Marking analysis: A learner attempts the following task: “Extension practice (HL content beyond the prescribed SL syllabus): Outline the level of protein structure at which hydrogen bonding creates an alpha helix or beta pleated sheet.” Their response addresses only this point: “States that this is the secondary structure of a protein.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [2 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that this is the secondary structure of a protein.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that hydrogen bonds form between the carbonyl (C=O) and amine (N–H) groups of the polypeptide backbone, creating a regular, repeating shape.

    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. 17.

    Explain how the alveoli of the human lung are structurally adapted for efficient gas exchange. Refer to surface area, diffusion distance and blood supply.

    [4 marks] · no calculator

    Worked explanation

    1. Many small alveoli together create a large total exchange area, allowing many gas molecules to cross simultaneously.
    2. Thin alveolar epithelium and capillary endothelium keep the air-blood diffusion path short.
    3. Continuous capillary flow brings oxygen-poor blood and carries oxygenated blood away. Along with ventilation, this maintains partial-pressure gradients driving passive exchange.

    Common mistake: A rich blood supply helps by maintaining gradients, not by actively pumping oxygen through the alveolar wall.

    Marking points

    • States that alveoli provide a very large total surface area for diffusion.
    • States that the alveolar wall is one cell thick, giving a short diffusion distance.
    • States that alveoli are surrounded by a dense capillary network.
    • Links continuous blood flow to maintaining a steep concentration gradient for oxygen and carbon dioxide.

    Examiner tip: Each structural feature should be explicitly linked to how it increases the rate of diffusion.

18. 18.

    Marking analysis: A learner attempts the following task: “Explain how the alveoli of the human lung are structurally adapted for efficient gas exchange. Refer to surface area, diffusion distance and blood supply.” Their response addresses only this point: “States that alveoli provide a very large total surface area for diffusion.” 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 alveoli provide a very large total surface area for diffusion.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the alveolar wall is one cell thick, giving a short diffusion distance.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that alveoli are surrounded by a dense capillary network.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Links continuous blood flow to maintaining a steep concentration gradient for oxygen and carbon dioxide.

    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. 19.

    Extension practice (includes HL content beyond the prescribed SL syllabus): Outline the roles of villi and microvilli in increasing the efficiency of absorption in the small intestine.

    [3 marks] · no calculator

    Course scope: Optional enrichment

    Worked explanation

    1. Villi project from the intestinal lining, while microvilli project from the apical surface of each epithelial cell. They add area at two different scales.
    2. More membrane area accommodates more transport routes for digested nutrients, increasing potential absorption per length of intestine.
    3. Absorbed substances enter blood or lymph according to their processing and transport route; extra area does not imply that every nutrient simply diffuses into blood.

    Common mistake: Villi and microvilli are not interchangeable names for the same projection: microvilli belong to individual cells.

    Marking points

    • States that villi are finger-like projections of the intestinal wall, and microvilli are even smaller projections on the surface of the epithelial cells covering each villus.
    • States that both structures greatly increase the surface area of the small intestine available for the absorption of digested nutrients.
    • States that a greater surface area allows a greater rate of nutrient absorption into the blood or lymph.

    Examiner tip: Extension practice: General surface-area reasoning is shared content (B2.3.6), but the required microvillus adaptation is AHL. Guide references: B2.3.7. This item is enrichment, not evidence of prescribed SL exam coverage. Villi and microvilli are a classic example of the general biological principle that increased surface area increases the rate of exchange processes.

20. 20.

    Marking analysis: A learner attempts the following task: “Extension practice (includes HL content beyond the prescribed SL syllabus): Outline the roles of villi and microvilli in increasing the efficiency of absorption in the small intestine.” Their response addresses only this point: “States that villi are finger-like projections of the intestinal wall, and microvilli are even smaller projections on the surface of the epithelial cells covering each villus.” 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 villi are finger-like projections of the intestinal wall, and microvilli are even smaller projections on the surface of the epithelial cells covering each villus.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that both structures greatly increase the surface area of the small intestine available for the absorption of digested nutrients.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a greater surface area allows a greater rate of nutrient absorption into the blood or lymph.

    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. 21.

    Explain gas exchange between alveolar air and blood in the lungs, referring to diffusion driven by partial-pressure differences.

    [4 marks] · no calculator

    Worked explanation

    1. Oxygen partial pressure is higher in alveolar air than in arriving venous blood, so oxygen diffuses into blood across the thin air-blood barrier.
    2. Carbon dioxide has the opposite partial-pressure difference and diffuses from blood into the alveolus.
    3. Ventilation renews alveolar air and circulation renews blood, preventing rapid equilibration from stopping net exchange.

    Common mistake: Comparing total gas concentrations across air and blood can mislead; partial-pressure differences specify the driving force across the interface.

    Marking points

    • States that oxygen partial pressure is higher in alveolar air than in the venous blood arriving at the lungs, so oxygen diffuses into the blood.
    • States that carbon dioxide partial pressure is higher in arriving blood than in alveolar air, so carbon dioxide diffuses into the alveoli.
    • States that gas exchange occurs across the thin alveolar and capillary walls (a short diffusion distance), which speeds up the rate of exchange.
    • States that blood flow and ventilation maintain the partial-pressure differences driving diffusion.

    Examiner tip: Across the air-blood interface, compare gas partial pressures rather than total gas concentrations in two different media. Diffusion is passive, although ventilation and circulation require energy.

22. 22.

    Marking analysis: A learner attempts the following task: “Explain gas exchange between alveolar air and blood in the lungs, referring to diffusion driven by partial-pressure differences.” Their response addresses only this point: “States that oxygen partial pressure is higher in alveolar air than in the venous blood arriving at the lungs, so oxygen diffuses into the blood.” 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 oxygen partial pressure is higher in alveolar air than in the venous blood arriving at the lungs, so oxygen diffuses into the blood.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that carbon dioxide partial pressure is higher in arriving blood than in alveolar air, so carbon dioxide diffuses into the alveoli.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that gas exchange occurs across the thin alveolar and capillary walls (a short diffusion distance), which speeds up the rate of exchange.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that blood flow and ventilation maintain the partial-pressure differences driving diffusion.

    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. 23.

    Extension practice (includes HL content beyond the prescribed SL syllabus): Outline the roles of the SA node and AV node in coordinating the cardiac cycle.

    [3 marks] · no calculator

    Course scope: Optional enrichment

    Worked explanation

    1. The SA node in the right atrium spontaneously initiates an electrical wave that spreads through the atria, causing atrial contraction.
    2. The AV node delays onward conduction, allowing blood to move into the ventricles before they contract.
    3. Conduction then reaches ventricular muscle through the specialised conducting pathway, coordinating ventricular ejection after filling.

    Common mistake: Simultaneous atrial and ventricular contraction would impair filling; the AV delay is functional, not an accidental pause.

    Marking points

    • States that the sinoatrial (SA) node, located in the right atrium, acts as the heart's natural pacemaker, generating an electrical impulse that triggers atrial contraction.
    • States that the impulse spreads across both atria, causing them to contract, and reaches the atrioventricular (AV) node.
    • States that the AV node briefly delays the impulse before passing it on to the ventricles, ensuring the atria finish contracting and emptying before the ventricles contract.

    Examiner tip: Extension practice: Detailed pacemaker-to-cardiac-cycle coordination is AHL; general feedback control of heart rate is separately shared content (C3.1.14). Guide references: B3.2.15-B3.2.16. This item is enrichment, not evidence of prescribed SL exam coverage. The AV node's delay is essential: it prevents the atria and ventricles from contracting simultaneously, which would prevent efficient filling and pumping of blood.

24. 24.

    Marking analysis: A learner attempts the following task: “Extension practice (includes HL content beyond the prescribed SL syllabus): Outline the roles of the SA node and AV node in coordinating the cardiac cycle.” Their response addresses only this point: “States that the sinoatrial (SA) node, located in the right atrium, acts as the heart's natural pacemaker, generating an electrical impulse that triggers atrial contraction.” 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 the sinoatrial (SA) node, located in the right atrium, acts as the heart's natural pacemaker, generating an electrical impulse that triggers atrial contraction.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the impulse spreads across both atria, causing them to contract, and reaches the atrioventricular (AV) node.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the AV node briefly delays the impulse before passing it on to the ventricles, ensuring the atria finish contracting and emptying before the ventricles contract.

    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. 25.

    Extension practice (HL content beyond the prescribed SL syllabus): Explain how the structure of the nephron enables both filtration and selective reabsorption during urine formation.

    [3 marks] · no calculator

    Course scope: Optional enrichment

    Worked explanation

    1. High pressure in glomerular capillaries drives water and small solutes into Bowman's space through the filtration barrier; cells and most plasma proteins remain in blood.
    2. The proximal tubule returns useful solutes such as glucose to blood through selective membrane transport. Water follows osmotically, so filtrate is not simply all excreted.
    3. After further reabsorption and secretion, remaining tubular fluid becomes urine containing urea and variable water/ions. Some urea is reabsorbed too.

    Common mistake: Glomerular filtration does not actively choose wastes; selective recovery occurs later along the tubule.

    Marking points

    • States that in the glomerulus, high blood pressure forces water and small solutes (e.g. glucose, salts, urea) out of the capillaries and into the Bowman's capsule, while blood cells and large proteins are retained (ultrafiltration).
    • States that as filtrate passes along the nephron tubule (especially the proximal convoluted tubule), useful substances such as glucose and some salts are selectively reabsorbed back into the blood by active transport and diffusion.
    • States that the water and solutes remaining in tubular fluid after reabsorption and secretion form urine, which contains urea and variable amounts of water and ions; some filtered urea is reabsorbed.

    Examiner tip: Extension practice: Kidney excretion, glomerular filtration and proximal-tubule reabsorption are AHL. Guide references: D3.3.7-D3.3.8. This item is enrichment, not evidence of prescribed SL exam coverage. Glomerular filtration is pressure-driven, with a barrier that discriminates by size and charge. Tubular reabsorption is selective and includes passive as well as energy-dependent transport.

26. 26.

    Marking analysis: A learner attempts the following task: “Extension practice (HL content beyond the prescribed SL syllabus): Explain how the structure of the nephron enables both filtration and selective reabsorption during urine formation.” Their response addresses only this point: “States that in the glomerulus, high blood pressure forces water and small solutes (e.g. glucose, salts, urea) out of the capillaries and into the Bowman's capsule, while blood cells and large proteins are retained (ultrafiltration).” 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 in the glomerulus, high blood pressure forces water and small solutes (e.g. glucose, salts, urea) out of the capillaries and into the Bowman's capsule, while blood cells and large proteins are retained (ultrafiltration).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that as filtrate passes along the nephron tubule (especially the proximal convoluted tubule), useful substances such as glucose and some salts are selectively reabsorbed back into the blood by active transport and diffusion.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the water and solutes remaining in tubular fluid after reabsorption and secretion form urine, which contains urea and variable amounts of water and ions; some filtered urea is reabsorbed.

    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. 27.

    Extension practice (includes HL content beyond the prescribed SL syllabus): Explain the role of haemoglobin in the transport of oxygen in the blood, referring to its structure.

    [3 marks] · no calculator

    Course scope: Optional enrichment

    Worked explanation

    1. Haemoglobin has four polypeptide subunits, each with an iron-containing haem group, allowing one molecule to bind up to four oxygen molecules.
    2. Binding is reversible: high oxygen partial pressure in the lungs favours loading as oxyhaemoglobin.
    3. Lower oxygen partial pressure in respiring tissues favours unloading, supplying oxygen for aerobic respiration rather than trapping it permanently on the carrier.

    Common mistake: Oxygen is reversibly bound to haem iron, not irreversibly incorporated into the globin chain.

    Marking points

    • States that haemoglobin is a globular protein found in red blood cells, made of four polypeptide subunits, each containing a haem group with an iron ion.
    • States that oxygen binds reversibly to the iron ion in each haem group, forming oxyhaemoglobin, in regions of high oxygen concentration such as the lungs.
    • States that haemoglobin releases oxygen in regions of lower oxygen concentration, such as respiring tissues, where it is needed.

    Examiner tip: Extension practice: The required four-subunit, haem-containing haemoglobin structure is AHL, as are the guide's haemoglobin transport adaptations; general gas exchange is shared content. Guide references: B1.2.11; B3.1.11. This item is enrichment, not evidence of prescribed SL exam coverage. Haemoglobin's binding to oxygen is reversible, not permanent — this is exactly what allows it to load oxygen in the lungs and unload it again at respiring tissues elsewhere in the body.

28. 28.

    Marking analysis: A learner attempts the following task: “Extension practice (includes HL content beyond the prescribed SL syllabus): Explain the role of haemoglobin in the transport of oxygen in the blood, referring to its structure.” Their response addresses only this point: “States that haemoglobin is a globular protein found in red blood cells, made of four polypeptide subunits, each containing a haem group with an iron ion.” 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 haemoglobin is a globular protein found in red blood cells, made of four polypeptide subunits, each containing a haem group with an iron ion.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that oxygen binds reversibly to the iron ion in each haem group, forming oxyhaemoglobin, in regions of high oxygen concentration such as the lungs.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that haemoglobin releases oxygen in regions of lower oxygen concentration, such as respiring tissues, where it is needed.

    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. 29.

    Extension practice (HL content beyond the prescribed SL syllabus): Explain how antagonistic pairs of muscles, such as the biceps and triceps, bring about movement at a joint.

    [3 marks] · no calculator

    Course scope: Optional enrichment

    Syllabus placement evidence: B3.3: no SL content, printed p.60. Retain the existing HL-depth task as optional enrichment outside SL requirements. This corrects applicability and navigation, not original content or SL completion.

    Worked explanation

    1. A shortening skeletal muscle exerts a pull through its tendon; it cannot actively push the bone back.
    2. At the elbow, biceps contraction with triceps relaxation produces flexion. For extension, triceps contracts while biceps relaxes.
    3. Opposing pulls provide control in both directions about the joint. Several muscles, gravity and elastic recoil can also contribute in real movements.

    Common mistake: The relaxing biceps does not actively push the forearm straight; the opposing triceps provides the extending pull.

    Marking points

    • States that a contracting skeletal muscle pulls rather than pushes; opposing muscles can therefore produce movements in opposite directions at a joint.
    • States that when the biceps contracts (and the triceps relaxes), the arm bends (flexion) at the elbow.
    • States that when the triceps contracts (and the biceps relaxes), the arm straightens (extension) at the elbow.

    Examiner tip: Extension practice: Antagonistic muscle action and joint mechanics fall within AHL-only B3.3. Guide references: B3.3.3; B3.3.5-B3.3.6. This item is enrichment, not evidence of prescribed SL exam coverage. At the elbow, biceps and triceps provide opposing flexion and extension. Do not generalise this to a requirement that every skeletal movement has exactly one opposing muscle pair; gravity, elastic recoil and several cooperating muscles can contribute.

30. 30.

    Marking analysis: A learner attempts the following task: “Extension practice (HL content beyond the prescribed SL syllabus): Explain how antagonistic pairs of muscles, such as the biceps and triceps, bring about movement at a joint.” Their response addresses only this point: “States that a contracting skeletal muscle pulls rather than pushes; opposing muscles can therefore produce movements in opposite directions at a joint.” 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 a contracting skeletal muscle pulls rather than pushes; opposing muscles can therefore produce movements in opposite directions at a joint.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that when the biceps contracts (and the triceps relaxes), the arm bends (flexion) at the elbow.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that when the triceps contracts (and the biceps relaxes), the arm straightens (extension) at the elbow.

    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. 31.

    Explain, in terms of hydrogen bonding, why water has a high specific heat capacity, and explain one biological advantage this property provides to aquatic organisms.

    [3 marks] · no calculator

    Worked explanation

    1. Water molecules interact through a dynamic network of hydrogen bonds. Absorbed thermal energy partly disrupts/rearranges that network rather than all increasing molecular kinetic energy.
    2. A relatively large energy input is therefore needed for a small temperature rise: water has high specific heat capacity.
    3. An aquatic animal experiences slower habitat temperature swings than it would in an equivalent low-heat-capacity medium, reducing rapid disruption of enzyme activity and physiology.

    Common mistake: High heat capacity does not mean water cannot change temperature, only that more energy is needed per unit mass per degree.

    Marking points

    • States that water molecules form extensive hydrogen bonds with one another.
    • Explains that a large amount of thermal energy must be absorbed to break these hydrogen bonds and increase the kinetic energy (and therefore temperature) of the water molecules, giving water a high specific heat capacity.
    • Explains a valid biological advantage, such as aquatic environments (and the bodies of organisms, which are largely water) resisting rapid temperature fluctuations, providing a more thermally stable habitat/internal environment than air.

    Examiner tip: For high specific heat capacity, relate water's hydrogen-bond network to its ability to absorb energy with a relatively small temperature rise. Other properties need their own explanation: solvent behaviour, for example, depends on polarity and interactions with the particular solute.

32. 32.

    Marking analysis: A learner attempts the following task: “Explain, in terms of hydrogen bonding, why water has a high specific heat capacity, and explain one biological advantage this property provides to aquatic organisms.” Their response addresses only this point: “States that water molecules form extensive hydrogen bonds with one another.” 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 water molecules form extensive hydrogen bonds with one another.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that a large amount of thermal energy must be absorbed to break these hydrogen bonds and increase the kinetic energy (and therefore temperature) of the water molecules, giving water a high specific heat capacity.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains a valid biological advantage, such as aquatic environments (and the bodies of organisms, which are largely water) resisting rapid temperature fluctuations, providing a more thermally stable habitat/internal environment than air.

    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. 33.

    Explain how the structure of alveoli in the lungs is adapted to maximize the rate of gas exchange.

    [3 marks] · no calculator

    Worked explanation

    1. Dividing the lung into many alveoli creates a large combined surface for oxygen and carbon dioxide exchange.
    2. Each thin alveolar wall lies next to thin capillary endothelium, minimising diffusion distance.
    3. Perfusion removes absorbed oxygen and supplies carbon dioxide; ventilation renews air. Together these prevent the gas partial pressures from rapidly equalising.

    Common mistake: An alveolus is not efficient because it is large individually; many small alveoli supply large total area.

    Marking points

    • States that the very large number of small alveoli provides a large total surface area for gas diffusion.
    • States that alveolar walls are only one cell thick, minimizing the diffusion distance between the air in the alveolus and the blood in the surrounding capillaries.
    • States that alveoli are surrounded by a dense network of capillaries, maintaining a steep concentration gradient by continuously removing diffused oxygen and delivering carbon dioxide.

    Examiner tip: Every gas exchange surface adaptation across different organisms (alveoli, gills, leaf mesophyll) reduces to the same three principles: large surface area, short diffusion distance, and a maintained concentration gradient.

34. 34.

    Marking analysis: A learner attempts the following task: “Explain how the structure of alveoli in the lungs is adapted to maximize the rate of gas exchange.” Their response addresses only this point: “States that the very large number of small alveoli provides a large total surface area for gas diffusion.” 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 the very large number of small alveoli provides a large total surface area for gas diffusion.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that alveolar walls are only one cell thick, minimizing the diffusion distance between the air in the alveolus and the blood in the surrounding capillaries.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that alveoli are surrounded by a dense network of capillaries, maintaining a steep concentration gradient by continuously removing diffused oxygen and delivering carbon dioxide.

    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.

35. 35.

    Explain why increasing the temperature beyond an enzyme's optimum causes a sharp decrease in reaction rate, with reference to enzyme structure.

    [3 marks] · no calculator

    Worked explanation

    1. Above the observed optimum, heat increasingly destabilises interactions maintaining the enzyme's active conformation.
    2. The active site's shape and chemical arrangement change, so fewer productive enzyme-substrate complexes form even though molecular collisions are faster.
    3. Loss of active enzyme then outweighs the kinetic benefit of warming, causing the sharp rate decline. Severe denaturation often persists after cooling, unlike simple cold-induced slowing.

    Common mistake: Enzymes are molecules, not living cells; describing them as dead does not explain loss of active-site function.

    Marking points

    • States that an enzyme's active site has a specific three-dimensional shape, maintained by hydrogen bonds and other interactions within its tertiary structure.
    • Explains that excessive heat causes these bonds to break, altering (denaturing) the enzyme's tertiary structure and therefore the shape of its active site.
    • Explains that once the active site's shape no longer complements the substrate, the enzyme can no longer bind the substrate effectively, causing a sharp drop in reaction rate as more enzyme molecules denature.

    Examiner tip: Denaturation is irreversible under most conditions — this is why enzyme activity does not simply return when temperature is lowered back toward the optimum after severe overheating, unlike the reversible effect of temperatures only slightly below optimum.

36. 36.

    Marking analysis: A learner attempts the following task: “Explain why increasing the temperature beyond an enzyme's optimum causes a sharp decrease in reaction rate, with reference to enzyme structure.” Their response addresses only this point: “States that an enzyme's active site has a specific three-dimensional shape, maintained by hydrogen bonds and other interactions within its tertiary structure.” 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 an enzyme's active site has a specific three-dimensional shape, maintained by hydrogen bonds and other interactions within its tertiary structure.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that excessive heat causes these bonds to break, altering (denaturing) the enzyme's tertiary structure and therefore the shape of its active site.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that once the active site's shape no longer complements the substrate, the enzyme can no longer bind the substrate effectively, causing a sharp drop in reaction rate as more enzyme molecules denature.

    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.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.