Exchange and Transport — Diagnostic Tests

Unit Tests

UT-1: Haemoglobin and the Oxygen Dissociation Curve

Question:

The oxygen dissociation curve for adult haemoglobin is sigmoidal (S-shaped). Fetal haemoglobin has a dissociation curve that lies to the left of adult haemoglobin.

(a) Explain why the oxygen dissociation curve for adult haemoglobin is sigmoidal rather than a straight line.

(b) Describe and explain the Bohr effect and its significance for oxygen delivery to actively respiring tissues.

(c) Explain why the fetal haemoglobin dissociation curve lies to the left of the adult curve, and explain why this is physiologically important.

(d) At high altitude, the partial pressure of oxygen is lower. Over several weeks, the body increases its production of red blood cells. Explain the mechanism that triggers this increase and how it compensates for the lower oxygen partial pressure.

Solution:

(a) The curve is sigmoidal because of cooperative binding. Haemoglobin has four haem groups, each capable of binding one O$_2$ molecule. When the first O$_2$ molecule binds to one haem group, it induces a conformational change in the haemoglobin molecule (quaternary structure change) that increases the affinity of the remaining three haem groups for oxygen. This means the second O$_2$ molecule binds more easily than the first, the third more easily than the second, and so on. This positive cooperativity produces the characteristic S-shape: a slow initial rise (first O$_2$ binding is difficult at low pO$_2$), followed by a steep rise as cooperativity takes effect, and a plateau as all four binding sites become occupied.

(b) The Bohr effect describes the observation that an increase in CO$_2$ concentration (or a decrease in pH) causes the oxygen dissociation curve to shift to the right. In actively respiring tissues, CO$_2$ is produced as a waste product of respiration. CO$_2$ diffuses into red blood cells and is converted to $\text{HCO}_3^-$ and H$^+$ by carbonic anhydrase: $\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$. The increased H$^+$ concentration (lower pH) causes haemoglobin to change shape, reducing its affinity for oxygen and promoting oxygen unloading. A rightward shift means that at any given pO$_2$, haemoglobin is less saturated with oxygen — it releases oxygen more readily. This ensures that tissues with high metabolic rates (high CO$_2$ production) receive more oxygen, precisely where it is most needed.

(c) Fetal haemoglobin (HbF) has a higher affinity for oxygen than adult haemoglobin (HbA) at any given partial pressure of oxygen, so its dissociation curve lies to the left. This is because fetal haemoglobin has a different quaternary structure (it has two gamma chains instead of two beta chains), which reduces its sensitivity to the Bohr effect and increases its oxygen affinity. This is physiologically important because the fetus must extract oxygen from the mother's blood in the placenta. In the placenta, the pO$_2$ in fetal blood is lower than in maternal blood. If fetal haemoglobin had the same affinity as adult haemoglobin, it would not be able to load oxygen effectively from maternal blood. The higher affinity of HbF ensures that oxygen diffuses from maternal blood (where HbA releases it) into fetal blood (where HbF binds it).

(d) At high altitude, the lower pO$_2$ means that haemoglobin is less fully saturated with oxygen in the lungs, reducing oxygen delivery to tissues. The kidneys detect the lower oxygen partial pressure and respond by secreting the hormone erythropoietin (EPO). EPO stimulates the bone marrow to increase the production of red blood cells (erythropoiesis). More red blood cells means more haemoglobin in the blood, which increases the oxygen-carrying capacity of the blood. Although each haemoglobin molecule is still less saturated at altitude, the increased total haemoglobin concentration compensates by carrying more oxygen per unit volume of blood. Additionally, increased 2,3-BPG (2,3-bisphosphoglycerate) production over time shifts the dissociation curve slightly to the right, promoting oxygen unloading in the tissues.

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UT-2: Xylem Transport and Transpiration

Question:

Water movement through a plant involves both the apoplast and symplast pathways, and is driven by the transpiration stream.

(a) Describe the cohesion-tension theory as the mechanism for water movement through the xylem.

(b) Explain the difference between the apoplast and symplast pathways of water movement across the root cortex to the xylem.

(c) Describe the adaptations of xylem vessels for their function in transporting water, and explain why xylem vessels are dead at maturity.

(d) Explain how environmental factors (temperature, humidity, wind speed, and light intensity) affect the rate of transpiration.

Solution:

(a) The cohesion-tension theory explains water movement through the xylem as follows:

1. Transpiration — water evaporates from the spongy mesophyll cells inside the leaf through the stomata, creating a water potential gradient between the inside of the leaf and the outside atmosphere.
2. Tension — the evaporation of water from mesophyll cell walls creates a negative pressure (tension) that pulls water upwards through the xylem vessels from the roots to the leaves. Water molecules are pulled in a continuous column due to the strong cohesion (hydrogen bonding) between water molecules.
3. Adhesion — water molecules also adhere to the hydrophilic walls of the xylem vessels, helping to maintain the column against gravity.
4. Root pressure — contributes a small additional push (osmotic uptake of water from the soil into root xylem), but the primary driving force is the tension created by transpiration at the leaves.

The column of water in the xylem is under tension (negative pressure), and this tension pulls water upwards from the roots. The cohesion between water molecules prevents the column from breaking (up to a point — if tension exceeds cohesion, cavitation occurs, forming an air bubble that blocks the vessel).

(b) Apoplast pathway: water moves through the cell walls and intercellular spaces of the root cortex without crossing any cell membranes. It moves by diffusion along the water potential gradient. This pathway offers the least resistance to water flow and is the fastest route. However, it is blocked by the Casparian strip (a band of suberin in the cell walls of the endodermis), which is impermeable to water.

Symplast pathway: water enters the cytoplasm of root hair cells by osmosis and passes from cell to cell through plasmodesmata (cytoplasmic connections between adjacent cells). Water may also move through the vacuoles (vacuolar pathway, a sub-route of the symplast). This pathway is slower because water must cross cell membranes.

When the apoplast pathway is blocked by the Casparian strip at the endodermis, water is forced into the symplast pathway. This ensures that all water entering the xylem has passed through the selectively permeable cell membrane of the endodermal cells, allowing the plant to control the entry of mineral ions and solutes.

(c) Xylem vessel adaptations:

• Dead at maturity: the cell contents (cytoplasm, organelles) break down, leaving a hollow lumen that offers minimal resistance to water flow. Being dead means they do not require energy or nutrients.
• Lignified walls: the walls are impregnated with the waterproof polymer lignin, which provides mechanical strength to withstand the tension (negative pressure) of the water column and prevents the vessel walls from collapsing inward. Lignin also makes xylem vessels impermeable to water.
• Continuous tubes: xylem vessel elements are arranged end-to-end; the end walls break down completely (or are perforated by pits), forming continuous columns for uninterrupted water flow.
• Pits: thin regions of the cell wall without lignin allow lateral movement of water between adjacent vessels.

(d) Environmental effects on transpiration rate:

• Temperature: higher temperature increases the kinetic energy of water molecules, increasing the rate of evaporation from mesophyll cell surfaces and increasing transpiration. Higher temperature also increases the water vapour holding capacity of air.
• Humidity: lower humidity (drier air) increases the water potential gradient between the leaf air spaces and the external atmosphere, increasing the rate of diffusion of water vapour out of the leaf (higher transpiration). High humidity reduces this gradient, decreasing transpiration.
• Wind speed: higher wind speed removes the layer of water vapour that accumulates near the leaf surface (boundary layer), maintaining a steep water vapour concentration gradient and increasing transpiration. Still air allows the boundary layer to build up, reducing transpiration.
• Light intensity: higher light intensity causes stomata to open (guard cells take up K$^+$, water enters by osmosis, guard cells become turgid), increasing the area through which water vapour can diffuse out of the leaf, thereby increasing transpiration. In darkness, stomata close, reducing transpiration.

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UT-3: Mass Flow Hypothesis and Phloem Transport

Question:

The mass flow hypothesis (pressure flow hypothesis) explains the transport of organic solutes (primarily sucrose) in the phloem from source to sink.

(a) Describe the mass flow hypothesis for the translocation of sucrose in the phloem.

(b) Explain the evidence supporting the mass flow hypothesis and the evidence against it.

(c) Explain how aphids were used to provide evidence for the direction of flow in phloem, and describe what the results showed.

(d) Explain why phloem transport is described as an active process, whereas xylem transport is described as a passive process.

Solution:

(a) The mass flow hypothesis:

1. Loading at the source: sucrose is actively transported (by companion cells using ATP) from photosynthetic cells (e.g., mesophyll) into the phloem sieve tube elements. This lowers the water potential inside the sieve tubes.
2. Water uptake: water enters the sieve tubes by osmosis from the xylem (and surrounding cells), creating a high hydrostatic pressure at the source end.
3. Mass flow: the high hydrostatic pressure at the source pushes the sap (sucrose solution) through the phloem towards the sink (e.g., root, growing tip, storage organ). Sucrose is transported along with the flow of water.
4. Unloading at the sink: sucrose is removed from the sieve tubes at the sink (by active transport or diffusion into sink cells). This raises the water potential inside the sieve tubes.
5. Water exits: water leaves the sieve tubes by osmosis into the surrounding tissues (or xylem), reducing the hydrostatic pressure at the sink end.

The pressure gradient (high pressure at source, low pressure at sink) drives the bulk flow of sap through the phloem.

(b) Evidence supporting the mass flow hypothesis:

• Phloem sap has a relatively high sugar concentration (up to 30%), consistent with mass flow driven by osmotic gradients.
• There is a pressure gradient in the phloem: ringing experiments (removing a ring of bark including phloem) cause a bulge above the ring due to sap accumulation (high pressure at source, blocked flow), demonstrating that transport occurs in the phloem and that there is pressure.
• Aphid stylet experiments show phloem sap exudes under pressure when a sieve tube is punctured.
• Companion cells have many mitochondria, consistent with the ATP requirement for active loading of sucrose.

Evidence against the mass flow hypothesis:

• Sucrose is transported to all sinks, not just the sink with the lowest hydrostatic pressure (the model predicts flow driven by a single pressure gradient, but multiple sinks are supplied simultaneously).
• The speed of translocation (approximately 0.5-1 m per hour) is faster than would be expected from simple diffusion alone but slower than bulk flow in a rigid pipe, suggesting some resistance in the sieve tubes.
• Sieve plates (with pores) between sieve tube elements would create resistance to mass flow, yet translocation is still efficient.

(c) Aphids (e.g., pea aphids) feed by inserting their stylets (mouthparts) into phloem sieve tubes to feed on the sap. Researchers used this by anaesthetising the aphid and cutting its body off, leaving the stylet embedded in the phloem. Phloem sap exuded from the severed stylet, and the sap was collected and analysed. The sap contained a high concentration of sucrose and amino acids, confirming that phloem transports organic solutes. When stylets were inserted into phloem at different points along the plant (e.g., source leaf vs root), the sap always flowed away from the source (the exudation was directional), confirming that translocation is unidirectional from source to sink (at least in a given sieve tube). Radioactive labelling ($^{14}$C) of sucrose in a source leaf showed that the labelled sucrose appeared in the phloem and moved towards the sink, providing further confirmation of the direction of flow.

(d) Phloem transport is active because: (1) sucrose is actively loaded into sieve tube elements at the source by companion cells, using ATP (this is the energy-requiring step that creates the osmotic gradient); (2) the process can be inhibited by respiratory poisons (e.g., cyanide, which inhibits ATP production), which stop translocation. Xylem transport is passive because: (1) water movement is driven by the passive evaporation of water from leaves (transpiration), which creates a tension that pulls water up through the xylem — no metabolic energy is directly required for water movement through xylem vessels; (2) xylem vessels are dead cells with no metabolic activity. (Root pressure, a minor component of xylem transport, does involve active transport of ions, but the main driver — transpiration pull — is passive.)

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Integration Tests

IT-1: Gas Exchange Adaptations and Surface Area to Volume Ratio (with Cells)

Question:

Single-celled organisms such as amoeba rely on simple diffusion for gas exchange, whereas larger multicellular organisms require specialised gas exchange surfaces and mass transport systems.

(a) Explain the relationship between surface area to volume ratio and the need for specialised gas exchange systems in large organisms. Your answer should include a calculation.

(b) Describe the adaptations of the mammalian lung alveoli for efficient gas exchange.

(c) Explain how the structure of the alveolar epithelium relates to the structure and function of the capillary endothelium adjacent to it, with reference to the cell types involved.

(d) Describe the mechanism by which oxygen is loaded onto haemoglobin in the lungs and unloaded in the tissues. Include the concept of partial pressure in your answer.

Solution:

(a) As an organism increases in size, its volume (and therefore its metabolic demand, which is proportional to the number of cells) increases faster than its surface area. For a cube of side length $s$: surface area $= 6s^2$ and volume $= s^3$. The SA:V ratio $= \frac{6s^2}{s^3} = \frac{6}{s}$. As $s$ increases, the SA:V ratio decreases. A smaller SA:V ratio means there is less surface area available for gas exchange relative to the volume of tissue that needs oxygen and produces CO$_2$. For example, a cube with $s = 1$ has SA:V $= 6$; a cube with $s = 10$ has SA:V $= 0.6$. The larger organism cannot meet its gas exchange needs by diffusion alone across its body surface, because the rate of diffusion is proportional to surface area but the demand is proportional to volume. Therefore, large organisms need specialised gas exchange surfaces (with large surface areas, thin barriers, and steep concentration gradients) and mass transport systems (circulatory systems) to move gases between the exchange surface and the tissues.

(b) Adaptations of alveoli:

• Large surface area: millions of alveoli in each lung provide an enormous total surface area for gas exchange (approximately 70 m$^2$ in humans).
• Thin diffusion barrier: the alveolar epithelium is a single layer of squamous epithelial cells (type I pneumocytes); the capillary endothelium is also a single cell layer; the basement membranes of the two layers are fused — total diffusion distance is approximately 1 micrometre.
• Steep concentration gradient: maintained by continuous blood flow through capillaries (removing O$_2$-rich blood and bringing CO$_2$-rich blood) and ventilation (breathing in fresh air with high O$_2$ and low CO$_2$, breathing out stale air).
• Good blood supply: dense capillary network surrounding each alveolus maintains the concentration gradient.
• Surfactant: reduces surface tension, preventing alveolar collapse and maintaining surface area.

(c) The alveolar epithelium consists mainly of type I pneumocytes — extremely thin, flat squamous epithelial cells that provide a minimal diffusion distance for gases. They are so thin that the cytoplasm is barely visible under a light microscope. Between type I pneumocytes are type II pneumocytes, which secrete surfactant. Adjacent to the alveolar epithelium, the capillary endothelium is also a single layer of thin, flat endothelial cells. The basement membranes of the alveolar epithelium and capillary endothelium are fused together, meaning there is no interstitial space between them. This arrangement — thin epithelium + thin endothelium + fused basement membranes — minimises the total diffusion distance between the air in the alveolus and the blood in the capillary, maximising the rate of gas exchange. Both are specialised to be thin because they are not required to provide mechanical strength (the connective tissue around alveoli provides that).

(d) Loading in the lungs: In the alveoli, the partial pressure of oxygen ($p\text{O}_2$) is high (approximately 13 kPa) because fresh air is continuously ventilated in. The partial pressure of oxygen in the deoxygenated blood arriving at the lungs is low (approximately 5 kPa). This creates a concentration gradient: oxygen diffuses from the alveolar air across the alveolar epithelium and capillary endothelium into the blood, and then into red blood cells where it binds to haemoglobin. In the high $p\text{O}_2$ environment of the lungs, haemoglobin's affinity for oxygen is high (low CO$_2$ concentration means minimal Bohr effect), and the cooperative binding means that once the first O$_2$ binds, the remaining haem groups load rapidly. Haemoglobin becomes highly saturated (approximately 97-98%).

Unloading in the tissues: In respiring tissues, $p\text{O}_2$ is low because cells are continuously using oxygen for aerobic respiration. CO$_2$ concentration is high. The high CO$_2$ causes the Bohr effect — haemoglobin's affinity for oxygen decreases, and the dissociation curve shifts to the right. Haemoglobin releases oxygen, which diffuses from the blood into the tissues along the concentration gradient.

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IT-2: Digestion and Mass Transport in Animals (with Biological Molecules)

Question:

The small intestine is the main site of digestion and absorption of nutrients. Lipids present a particular challenge because they are insoluble in water.

(a) Describe the role of bile in the digestion and absorption of lipids.

(b) Explain how the products of lipid digestion are absorbed and transported in the body, including the formation of micelles, chylomicrons, and entry into the lymphatic system.

(c) The villi are supplied with a dense network of blood capillaries. Explain how glucose and amino acids absorbed from the small intestine are transported to the liver via the hepatic portal vein, and explain the significance of the hepatic portal system.

(d) Explain why the blood leaving the liver in the hepatic vein has a different composition from the blood arriving in the hepatic portal vein, with reference to the metabolic functions of the liver.

Solution:

(a) Bile is produced in the liver, stored in the gall bladder, and released into the small intestine (duodenum) via the bile duct. Bile contains bile salts, which are amphipathic molecules (having both hydrophilic and hydrophobic regions). Bile salts emulsify lipids — they break large fat globules into smaller droplets, increasing the surface area for the enzyme lipase to act upon. Emulsification is a physical process (not chemical digestion). Bile also neutralises the acidic chyme arriving from the stomach, raising the pH to the optimum for pancreatic lipase (approximately pH 7-8). Bile salts also form micelles that help transport the products of lipid digestion to the epithelial surface for absorption.

(b) Lipase breaks down triglycerides into monoglycerides, fatty acids, and glycerol. These products are hydrophobic and would clump together in the aqueous environment of the intestinal lumen. Instead, they are incorporated into micelles — small spherical aggregates of bile salts with their hydrophilic regions facing outwards and the lipid products in the hydrophobic centre. Micelles transport the lipid products to the surface of the intestinal epithelium. The lipid products diffuse across the epithelial cell membrane (they are lipid-soluble and can pass through the phospholipid bilayer by simple diffusion). Inside the epithelial cell, monoglycerides and fatty acids are re-esterified into triglycerides. These triglycerides, along with cholesterol and fat-soluble vitamins, are packaged into chylomicrons (lipoprotein particles with a protein coat). Chylomicrons are too large to enter blood capillaries, so they are released from the epithelial cell by exocytosis and enter the lacteal (a lymphatic vessel in the villus). They are transported via the lymphatic system and eventually enter the bloodstream at the thoracic duct (near the heart).

(c) Glucose and amino acids are absorbed from the small intestine into the blood capillaries of the villi. These capillaries drain into larger blood vessels that merge to form the hepatic portal vein, which carries blood directly from the digestive system to the liver. The hepatic portal system is significant because it ensures that all nutrients absorbed from the gut pass through the liver first, before entering the general systemic circulation. This allows the liver to process, store, or detoxify these nutrients before they reach the rest of the body. For example, the liver converts excess glucose to glycogen (glycogenesis) for storage, converts amino acids by deamination (removing the amino group and producing urea), and detoxifies alcohol and other harmful substances.

(d) The blood in the hepatic portal vein is rich in nutrients (glucose, amino acids, lipids) and potentially harmful substances absorbed from the gut. The liver processes these: (1) Glucose: excess glucose is converted to glycogen (glycogenesis); when blood glucose is low, glycogen is broken down to glucose (glycogenolysis). (2) Amino acids: excess amino acids are deaminated — the amino group is converted to ammonia (toxic), then to urea (less toxic) by the ornithine cycle; the remaining carbon skeleton is converted to carbohydrate or fat. (3) Detoxification: the liver breaks down alcohol, drugs, and other toxins. As a result, the blood leaving the liver in the hepatic vein has a lower concentration of glucose (regulated to approximately 90 mg per 100 cm$^3$), lower amino acid concentration, contains urea (for excretion by the kidneys), and has reduced levels of toxins. The liver also produces plasma proteins (e.g., fibrinogen, albumin) that enter the hepatic vein.

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IT-3: Plant Transport and Adaptations to Environment (with Ecology)

Question:

Xerophytic plants, such as marram grass, are adapted to reduce water loss in arid conditions. Their adaptations involve modifications to leaf structure, stomatal behaviour, and vascular tissue.

(a) Describe the adaptations of xerophytic leaves that reduce water loss.

(b) Explain how the rate of transpiration in a xerophyte differs from that in a typical mesophytic plant, and explain how this affects the rate of mineral ion uptake from the soil.

(c) In very hot conditions, some plants close their stomata to reduce water loss. Explain the consequence of this for the rate of photosynthesis, including the effect on the intracellular concentration of CO$_2$ and the Calvin cycle.

(d) Explain how the xylem structure in a xerophyte might differ from that of a mesophyte, and relate this to the need to withstand greater tension in the water column during drought conditions.

Solution:

(a) Xerophytic leaf adaptations to reduce water loss:

• Rolled leaves: the leaf curls inward, trapping a layer of moist air in the inner surface, reducing the water potential gradient between the leaf and the air.
• Sunken stomata: stomata are located in pits or grooves below the leaf surface, where they are sheltered from air currents and a layer of moist air accumulates, reducing transpiration.
• Thick waxy cuticle: a thick cuticle on the upper (and sometimes lower) leaf surface reduces water loss by cuticular transpiration.
• Reduced leaf surface area: small, needle-like leaves minimise the surface area from which water can evaporate.
• Hairs on the leaf surface: a dense layer of hairs traps moist air close to the leaf surface, reducing the water potential gradient.
• Fewer stomata: a lower density of stomata reduces the total area through which water vapour can diffuse.

(b) Xerophytes have a lower rate of transpiration than mesophytes due to their structural adaptations (rolled leaves, sunken stomata, thick cuticle, etc.). The transpiration stream (cohesion-tension mechanism) is therefore weaker, and the rate of water uptake from the soil is lower. Since mineral ions are transported through the xylem along with the water (carried in solution), the rate of mineral ion uptake and transport is also lower. However, xerophytes typically have deeper and more extensive root systems to access water from deeper soil layers, which partially compensates for the reduced transpiration-driven uptake at the root surface.

(c) When stomata close to conserve water: CO$_2$ can no longer diffuse into the leaf from the atmosphere, so the intracellular concentration of CO$_2$ in the mesophyll cells falls. CO$_2$ is a substrate for the enzyme Rubisco (ribulose bisphosphate carboxylase/oxygenase) in the Calvin cycle. With less CO$_2$ available, the rate of CO$_2$ fixation decreases, and the rate of the Calvin cycle slows. This reduces the production of triose phosphate (GP, GALP) and ultimately the production of glucose and other organic molecules. At very low CO$_2$ concentrations, Rubisco may act as an oxygenase (photorespiration), binding O$_2$ instead of CO$_2$, which is wasteful and further reduces photosynthetic efficiency. Therefore, there is a trade-off: closing stomata conserves water but reduces photosynthesis and limits growth.

(d) In xerophytes, the xylem vessels tend to have thicker, more heavily lignified walls compared to mesophytes. This is because during drought conditions, the rate of transpiration can fluctuate dramatically (e.g., very high during hot, windy days; very low at night). Rapid transpiration creates very high tension (negative pressure) in the water column within the xylem. Thicker lignification provides greater mechanical strength to resist the inward collapse of the vessel walls under this tension, preventing implosion. Additionally, xerophytes may have narrower xylem vessels, which are more resistant to cavitation (the formation of air bubbles when tension becomes too great) because the narrower bore produces a greater capillary action that helps maintain the water column. Some xerophytes also have tracheids (narrower, tapered xylem cells with pits) instead of or in addition to wide vessels, as tracheids are more resistant to cavitation.