Biology • Year 11 • Module 2 • Lesson 20
Autotrophs vs Heterotrophs, Full Synthesis
Apply the autotroph/heterotroph comparison to real data on gas exchange, transport medium composition, and day/night photosynthesis scenarios.
1. Interpret gas exchange data across a 24-hour period
A researcher measured the net gas exchange (in µmol/m²/s) of a potted tomato plant at three times during a 24-hour period. Positive values indicate net release; negative values indicate net uptake. 8 marks
| Time | Light conditions | Net O&sub2; (µmol/m²/s) | Net CO&sub2; (µmol/m²/s) |
|---|---|---|---|
| 10:00 am | Bright sunlight | +12.4 | −11.8 |
| 6:00 pm | Dim light (near sunset) | +0.2 | −0.1 |
| 2:00 am | Complete darkness | −4.6 | +4.3 |
1.1 At 10:00 am, both photosynthesis and cellular respiration are occurring simultaneously. Explain what the positive O&sub2; value tells you about the relative rates of these two processes at this time. 2 marks
1.2 At 6:00 pm, the net gas exchange values are close to zero. Identify the name for this condition and describe what is happening at the cellular level in terms of the two processes occurring. 2 marks
1.3 At 2:00 am, the plant is in darkness. A student claims: “At 2:00 am, the plant is behaving like an animal for gas exchange purposes.” Evaluate this claim using data from the table. 2 marks
1.4 The researcher also measured a mammal (rat) placed next to the plant under the same conditions. Predict the rat’s net O&sub2; and CO&sub2; values at 10:00 am and at 2:00 am. Explain your prediction. 2 marks
2. Interpret blood composition data, IQ3 application
The table below shows substance concentrations in blood sampled from five locations in a resting human. 7 marks
| Location | O&sub2; (units) | CO&sub2; (units) | Glucose (units) | Urea (units) |
|---|---|---|---|---|
| Pulmonary vein (leaving lungs) | 19 | 40 | 4.5 | 5.2 |
| Hepatic portal vein (gut→liver, post-meal) | 14 | 46 | 12.8 | 5.1 |
| Hepatic vein (leaving liver) | 13 | 48 | 4.6 | 7.9 |
| Renal vein (leaving kidneys) | 15 | 50 | 4.5 | 1.1 |
| Vena cava (returning to heart) | 12 | 52 | 4.2 | 1.3 |
2.1 Glucose rises sharply in the hepatic portal vein (4.5 → 12.8) but falls back to 4.6 in the hepatic vein. Name the organ responsible for this fall and explain the biological process involved. 2 marks
2.2 Urea rises from 5.1 to 7.9 across the liver, then falls from 7.9 to 1.1 across the kidneys. Identify which organ produces urea and which organ removes it, and explain the biological process for each. 3 marks
2.3 O&sub2; is highest in the pulmonary vein (19 units) and lowest in the vena cava (12 units). Explain this trend in terms of cellular activity in body tissues. 2 marks
3. Interpret xylem sap composition data
The table below shows xylem sap composition at two locations in a well-watered plant. 4 marks
| Substance | Root xylem (µmol/L) | Leaf petiole xylem (µmol/L) |
|---|---|---|
| NO&sub3;− | 2.4 | 1.1 |
| K+ | 3.8 | 2.0 |
| Ca2+ | 1.2 | 0.9 |
| Sucrose | trace | trace |
3.1 All mineral concentrations fall from root to leaf xylem. Explain why this occurs, referring to the function of leaf cells. 2 marks
3.2 Sucrose is present only in trace amounts in xylem sap, yet sucrose is described as the primary organic transport molecule in plants. Explain this apparent contradiction. Name the vessel you would sample to find high sucrose concentrations. 2 marks
4. Apply to a novel scenario, sealed glass terrarium
A student places a healthy fern plant and a live snail inside a sealed, transparent glass terrarium. The terrarium is left on a sunny windowsill for 24 hours, then moved into a dark cupboard for 24 hours. The student claims: “While the terrarium is on the windowsill, the snail has a guaranteed oxygen supply because the plant is photosynthesising.” 5 marks
4.1 Explain whether the student’s claim is scientifically accurate during the windowsill phase. Refer to the relative rates of photosynthesis and respiration for both organisms. 2 marks
4.2 Predict what will happen to O&sub2; levels in the sealed terrarium during the 24-hour dark phase. Explain your prediction by referring to the gas exchange processes occurring in both organisms. 2 marks
4.3 Identify which inquiry question (IQ1, IQ2, or IQ3) this scenario is primarily testing and justify your choice in one sentence. 1 mark
Q1.1, Relative rates at 10:00 am (2 marks)
The positive net O&sub2; value (+12.4) means photosynthesis is producing O&sub2; at a greater rate than cellular respiration is consuming it [1]. The net CO&sub2; value (−11.8) confirms photosynthesis is fixing more CO&sub2; than respiration releases [1]. Both processes are occurring simultaneously; photosynthesis simply dominates.
Marking criteria: 1 mark for explicitly stating photosynthesis rate > respiration rate (not just “photosynthesis is occurring”); 1 mark for linking to why the net value is positive (O&sub2; produced exceeds O&sub2; consumed).
Q1.2, Near-zero values at 6:00 pm (2 marks)
This is the light compensation point [1], the light intensity at which the rate of photosynthesis exactly equals the rate of cellular respiration. O&sub2; is produced by photosynthesis at the same rate as it is consumed by respiration, and CO&sub2; produced by respiration is consumed by photosynthesis at the same rate, so no net gas exchange is detectable [1].
Marking criteria: 1 mark for naming the compensation point; 1 mark for explaining equal rates of both processes results in no net gas exchange.
Q1.3, “Behaving like an animal” claim (2 marks)
The claim is substantially correct [1]. At 2:00 am in complete darkness, the plant’s net O&sub2; is negative (−4.6), it is consuming O&sub2;, and its net CO&sub2; is positive (+4.3), it is releasing CO&sub2;. This is exactly the same pattern as a heterotroph performing cellular respiration only. Photosynthesis has ceased because there is no light energy; only respiration operates [1]. (Note: the plant is not literally “becoming” an animal; it is behaving similarly for gas exchange only.)
Marking criteria: 1 mark for evaluating the claim as broadly correct; 1 mark for supporting with specific data values from the table (O&sub2; negative and/or CO&sub2; positive at night).
Q1.4, Rat gas exchange prediction (2 marks)
The rat’s net O&sub2; would be negative (consuming O&sub2;) and net CO&sub2; would be positive (releasing CO&sub2;) at both times, approximately the same magnitude regardless of light conditions [1]. This is because mammals are heterotrophs that perform only cellular respiration; they have no photosynthesis apparatus, so light does not change their gas exchange pattern. The rat’s values would be roughly similar to the plant’s 2:00 am values [1].
Marking criteria: 1 mark for predicting negative O&sub2; / positive CO&sub2; at both times; 1 mark for explaining that heterotrophs perform only cellular respiration and light has no effect on their gas exchange.
Q2.1, Glucose fall across the liver (2 marks)
The organ responsible is the liver [1]. After a meal, absorbed glucose floods the hepatic portal vein at high concentrations (12.8 units). The liver performs glycogenesisconverting excess glucose to glycogen for storage, bringing blood glucose back to the homeostatic set point (~4.5 units) before blood enters the general circulation. This prevents hyperglycaemia [1].
Marking criteria: 1 mark for identifying the liver; 1 mark for naming glycogenesis or describing glycogen storage as the mechanism that removes excess glucose.
Q2.2, Urea production and removal (3 marks)
Urea is produced by the liver [1]. The liver deaminates excess amino acids (amino group removed), releasing toxic ammonia (NH&sub3;), which the liver converts to the less toxic urea via the urea cycle [1]. Urea is removed by the kidneys, which filter urea from blood into the filtrate and excrete it in urine; this is why urea concentration falls from 7.9 to 1.1 across the kidneys [1].
Marking criteria: 1 mark for identifying liver as urea producer; 1 mark for explaining deamination of amino acids / urea cycle; 1 mark for identifying kidneys as the site of urea removal via filtration and excretion in urine.
Q2.3, O&sub2; trend from lungs to vena cava (2 marks)
O&sub2; falls from 19 to 12 units because all metabolically active body tissues extract O&sub2; from the blood via capillary exchange for use in cellular respiration [1]. As blood passes through the systemic capillary beds, O&sub2; diffuses down its partial pressure gradient from the blood into cells; the cumulative extraction by all tissues produces the overall fall visible in the vena cava [1].
Marking criteria: 1 mark for identifying that tissues consume O&sub2; for cellular respiration; 1 mark for explaining diffusion down partial pressure gradients in capillary beds as the mechanism.
Q3.1, Mineral fall from root to leaf (2 marks)
Mineral concentrations fall because leaf cells actively absorb minerals from the xylem sap as it passes through the leaf [1]. For example, NO&sub3;− is absorbed for amino acid synthesis; K+ is used for guard cell osmosis and enzyme activation; Ca2+ is used in cell wall construction. As xylem sap moves upward, it is progressively depleted of these minerals by the surrounding cells, lowering concentrations by the time the sap reaches the petiole [1].
Marking criteria: 1 mark for stating that leaf cells absorb minerals from xylem sap; 1 mark for providing a specific named function of at least one mineral, or explaining progressive depletion along the path.
Q3.2, Sucrose in xylem vs phloem (2 marks)
Sucrose is transported in phloem, not xylem. Xylem carries water and inorganic minerals (it does not transport organic sugars in significant quantities) [1]. Any trace sucrose in xylem may result from minor symplast leakage or sampling contamination. To find high sucrose concentrations, you would sample phloem sap, particularly near source leaves where sucrose is actively loaded [1].
Marking criteria: 1 mark for explaining xylem transports water + minerals, not organic sugars; 1 mark for correctly identifying phloem as the vessel to sample for sucrose.
Q4.1, Terrarium windowsill claim (2 marks)
The claim is broadly correct but overstated. During bright daylight, the fern’s photosynthesis rate exceeds its respiration rate, producing a net surplus of O&sub2; that is available to the snail [1]. However, the claim of a “guaranteed” supply depends on light intensity remaining above the compensation point; if sunlight dims (e.g. cloud cover) or the fern’s photosynthesis rate drops below both organisms’ combined respiration rates, O&sub2; would no longer be in net surplus [1].
Marking criteria: 1 mark for confirming the plant produces net O&sub2; during the light phase; 1 mark for qualifying the claim, it depends on photosynthesis rate remaining above the combined respiration rates.
Q4.2, Dark phase O&sub2; prediction (2 marks)
O&sub2; levels will fall continuously during the dark phase [1]. In darkness the fern cannot photosynthesise (no light energy), so it only performs cellular respiration, consuming O&sub2; and releasing CO&sub2;. The snail also respires continuously. Both organisms are now net consumers of O&sub2; in a sealed space, causing O&sub2; levels to drop and CO&sub2; levels to rise. The sealed system means no gas exchange with the outside air [1].
Marking criteria: 1 mark for predicting O&sub2; fall; 1 mark for explaining both organisms respire in darkness with no photosynthesis to replenish O&sub2; in the sealed system.
Q4.3, IQ identification (1 mark)
This scenario primarily tests IQ2the difference in nutrient and gas requirements between autotrophs and heterotrophs, because it contrasts the gas exchange behaviour of a photosynthetic autotroph (fern) and a heterotroph (snail) under different light conditions.
Marking criteria: 1 mark for IQ2 with a valid justification. Accept IQ3 with strong justification about how the sealed environment demonstrates composition changes in a shared air space.