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IB Biology: Photosynthesis and Limiting Factors (Paper 3: challenge)
IB Biology: Photosynthesis and Limiting Factors (Paper 3: challenge)
IB biology photosynthesis paper three, challenge: chemiosmosis, photorespiration, CAM adaptation, ATP and NADPH stoichiometry and energy efficiency.
Direct answers Answers in order: ATP, photorespiration, less daytime water loss, 18 ATP and 12 NADPH, 264 g, 12 units, carbon dioxide is no longer limiting, 120 kg, 7.02 per cent, and the Calvin cycle needs carbon dioxide.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: IB Biology: Photosynthesis and Limiting Factors |
| Level | IB |
02 Questions (10 items)
- FoundationWhat is produced when protons flow back through ATP synthase?
- FoundationIn hot dry conditions stomata close and leaf carbon dioxide falls. Which reaction does rubisco favour then?
- FoundationCAM plants open their stomata at night to fix carbon dioxide. What advantage does this give?
- IntermediateThe Calvin cycle uses 3 ATP and 2 NADPH per carbon dioxide fixed. What is needed for six molecules?
- IntermediateSplitting water releases 6 mol of oxygen. How many grams of carbon dioxide can be fixed at the same time, taking 44 g/mol?
- IntermediateA chloroplast suspension diluted to one fifth gives a rate of 2.4 units. What was the original rate?
- ChallengeAt one light intensity the 0.1 per cent carbon dioxide group outruns the 0.04 per cent group, but the curves merge at high light. What does this show?
- ChallengeA canopy fixes 12 g of glucose per square metre per day. What is that per hectare of 10 000 square metres, in kilograms?
- ChallengeA leaf absorbs 20 000 kJ of light and stores 90 g of glucose at 15.6 kJ per gram. What is the efficiency to two decimal places?
- Error redoA student argues that because the light reactions need no carbon dioxide, photosynthesis as a whole needs none. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | ATP | Chemiosmosis converts the proton motive force into chemical energy during photophosphorylation. |
| 2 | Photorespiration | Rubisco also has oxygenase activity, so oxygen competes with carbon dioxide at the same active site. |
| 3 | It cuts water loss during the day | Nights are cooler and damper, and the stored carbon still feeds the Calvin cycle while stomata stay shut by day. |
| 4 | 18 ATP and 12 NADPH | Six times 3 is 18 and six times 2 is 12. |
| 5 | 264 g | Six moles times 44 g/mol gives 264 g. |
| 6 | 12 units | Diluting to one fifth leaves a fifth of the activity, so 2.4 times 5 gives 12. |
| 7 | At high light carbon dioxide is no longer limiting, so both groups are held by the same other factor | Merging means one factor governs both, usually light intensity or temperature. |
| 8 | 120 kg | Twelve times 10 000 is 120 000 g, which is 120 kg. |
| 9 | 7.02 per cent | Ninety times 15.6 is 1 404 kJ, divided by 20 000 and times 100 gives 7.02. |
| 10 | The Calvin cycle needs carbon dioxide as its raw material, so without it no carbon can be fixed | The stages divide the work, and with no carbon dioxide there is simply no carbon to fix. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
The light reactions make ATP and NADPH in the thylakoid membrane while the Calvin cycle fixes carbon dioxide in the stroma, and the slowest of light, carbon dioxide and temperature sets the overall rate.Tutor Registration →
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