IB DP Biology: Photosynthesis: pigments, limiting factors, light-dependent reactions and the Calvin cycle (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover IB Diploma Programme Biology topic C1.3 (Photosynthesis), from the IB Biology guide, first assessment 2025. Questions 1 to 8 are SL and HL. Questions 9 to 12 test additional higher level statements and are labelled “HL only”. Marks are in bold brackets.

Study first with the study guide and revision notes. Course pages: IB DP Biology hub, printable checklist and free 10-minute diagnostics.

Questions

1. Write the word equation for photosynthesis and state the source of the oxygen released. [2]

2. A leaf extract was separated by paper chromatography. The solvent front moved 6.4 cm. Four spots were found: A (orange-yellow) at 6.1 cm, B (yellow) at 4.4 cm, C (blue-green) at 3.3 cm and D (yellow-green) at 2.2 cm. In the same system, reference Rf values are carotene 0.95, chlorophyll a 0.52 and chlorophyll b 0.34.

(a) Calculate the Rf value of B. [1] (b) Identify pigment D, using both colour and Rf. [2] (c) Explain why the start line was drawn in pencil. [1]

3. The absorbance of a pigment extract was measured at several wavelengths.

Wavelength / nm	400	430	470	520	560	620	660	700
Absorbance	0.55	0.90	0.35	0.05	0.04	0.25	0.70	0.10

(a) State the wavelength of greatest absorption. [1] (b) Explain why leaves containing this pigment look green. [2] (c) Outline what happens inside a pigment molecule when it absorbs a photon. [1]

4. A leaf was sealed in a chamber and lit for 20 minutes with light of one wavelength at a time. The carbon dioxide concentration started at 400 ppm each time.

Wavelength / nm	440	520	600	660	720
Final concentration / ppm	310	372	350	318	396

(a) Calculate the rate of carbon dioxide uptake at 520 nm, in ppm per minute. [1] (b) Calculate the rate at 660 nm as a percentage of that at 440 nm (nearest whole number). [2] (c) Describe the action spectrum these data would give. [2] (d) Suggest why photosynthesis still occurs at 520 nm. [1]

5. Compare absorption spectra and action spectra. [3]

6. A student plans a bubble-counting experiment on an aquatic plant, to see how light intensity affects its photosynthesis rate.

(a) State a hypothesis. [1] (b) Identify the independent and the dependent variable. [2] (c) Identify two variables to control and state how each is controlled. [2] (d) State how light intensity can be varied. [1]

7. The table shows bubbles released per minute by an aquatic plant at constant light intensity and carbon dioxide concentration.

Temperature / °C	10	20	30	40	50
Bubbles per minute	4	9	17	21	3

(a) Calculate the percentage increase in rate as the temperature goes from 20 °C to 30 °C. [1] (b) Explain the change in rate from 40 °C to 50 °C. [2] (c) At 10 °C, raising the light intensity gave almost no increase in rate. Explain this. [2]

8. A field crop was grown in open-air plots at ambient carbon dioxide and in plots enriched using FACE. Mean grain yield was 5.8 t ha⁻¹ in ambient plots and 6.5 t ha⁻¹ in enriched plots. The mean photosynthetic rate was 19.8 and 24.6 µmol m⁻² s⁻¹ respectively.

(a) Calculate the percentage change in grain yield. [1] (b) Calculate the percentage change in photosynthetic rate. [1] (c) Suggest why the change in yield is smaller than the change in rate. [1] (d) State one advantage of FACE over an enclosed greenhouse experiment. [1] (e) State one limitation of using these results to predict future yields. [1] (f) State one variable to control between plots. [1]

9. (HL only)

(a) Describe the structure of a photosystem. [2] (b) Explain the advantage of this structure compared with a single chlorophyll molecule. [2]

10. (HL only)

(a) State where photolysis occurs and name its three products. [2] (b) Explain how a proton gradient is formed and used to make ATP in a thylakoid. [4] (c) Outline how NADP is reduced. [2]

11. (HL only) In a chloroplast, 18 carbon dioxide molecules were fixed by the Calvin cycle. Assume each GP converted to TP uses 1 ATP and 1 reduced NADP, and each RuBP regenerated uses 1 ATP.

(a) State the number of GP molecules formed. [1] (b) State how many TP molecules are used to regenerate RuBP. [1] (c) State how many RuBP molecules are regenerated. [1] (d) Calculate how many glucose molecules could be made from the remaining TP. [1] (e) Calculate the total ATP and reduced NADP used. [2] (f) State the fraction of TP converted back to RuBP. [1]

12. (HL only) Algae photosynthesising steadily were moved to darkness. Relative amounts of RuBP and GP were measured.

Time after darkness / s	0	10	20	30
RuBP	100	62	35	20
GP	100	135	160	170

(a) Calculate the percentage change in each compound by 20 seconds. [2] (b) Explain these changes. [3] (c) Explain why removing carbon dioxide would stop photosystem II working. [3]

Answers

1. Carbon dioxide + water → glucose + oxygen [1]; the oxygen comes from the splitting of water [1]. [2] Examiner insight: Name reactants and products in full; saying the oxygen comes from carbon dioxide is a frequent error.

2. (a) 4.4 ÷ 6.4 = 0.69 [1] (b) D has Rf 2.2 ÷ 6.4 = 0.34 [1], matching chlorophyll b, and it is yellow-green; so D is chlorophyll b [1]. (c) Pencil is insoluble in the solvent, so it stays put and gives no extra spots [1]. [4] Examiner insight: Show the division and use both colour and Rf, as the question asks.

3. (a) 430 nm [1] (b) Absorbance is low at about 520 to 560 nm, so green light is not absorbed [1]; it is reflected or transmitted to the eye [1]. (c) An electron moves up to a higher energy level, becoming excited [1]. [4] Examiner insight: Give a wavelength with units, and link the green wavelengths to their low absorbance.

4. (a) (400 − 372) ÷ 20 = 1.4 ppm min⁻¹ [1] (b) Rate at 660 = (400 − 318) ÷ 20 = 4.1 [1]; rate at 440 = 90 ÷ 20 = 4.5; 4.1 ÷ 4.5 × 100 = 91% [1] (c) Two peaks, in blue-violet (440 nm, 4.5 ppm min⁻¹) and red (660 nm, 4.1) [1]; low rate in green (520 nm, 1.4) and almost none at 720 nm (0.2) [1]. (d) Accessory pigments absorb some green light and pass energy on [1]. [6] Examiner insight: Quote values from the table and name the regions of the spectrum.

5. Similarity: both show high values in blue-violet and red and low values in green [1]. Difference: an absorption spectrum records the light a pigment takes in [1], whereas an action spectrum shows the rate of photosynthesis against wavelength [1]. [3] Examiner insight: A difference needs both sides stated.

6. (a) If light intensity increases, the rate of photosynthesis (bubbles per minute) will increase, because more photons are absorbed [1]. (b) Independent: light intensity [1]. Dependent: bubbles per minute (rate of oxygen release) [1]. (c) Temperature, controlled by placing a water-filled tank between lamp and plant or using a water bath [1]; carbon dioxide concentration, controlled by using the same hydrogencarbonate solution [1]. (d) Change the distance between lamp and plant [1]. [6] Examiner insight: Give the method of control as well as the name of the variable.

7. (a) (17 − 9) ÷ 9 × 100 = 89% (88.9) [1] (b) Enzymes, including Rubisco, denature at high temperature [1]; the active site shape changes, so less carbon is fixed and the rate falls sharply [1]. (c) At 10 °C, enzyme activity is too low, so temperature is the limiting factor [1]; more light cannot raise the rate because the Calvin cycle cannot use the extra ATP and reduced NADP faster [1]. [5] Examiner insight: Name the limiting factor and say why more light has no effect.

8. (a) (6.5 − 5.8) ÷ 5.8 × 100 = 12.1% [1] (b) (24.6 − 19.8) ÷ 19.8 × 100 = 24.2% [1] (c) Other factors, such as mineral nutrients or water, limit growth, or carbon is used in respiration or in parts not harvested [1]. (d) Natural open-air conditions, so results apply better to real ecosystems [1]. (e) Other future changes, such as higher temperature, are not included in the experiment [1]. (f) Species or variety, soil type, watering or plot size [1]. [6] Examiner insight: Percentage change divides by the original value; name a specific controlled variable.

9. (a) A group of chlorophyll and accessory pigment molecules arranged in an array in a membrane [1]; a special chlorophyll at the centre is the reaction centre, where an excited electron is given off [1]. (b) Different pigments absorb different wavelengths, so more light is captured [1]; the energy is passed to the reaction centre, which a single molecule could not use to produce an emitted electron [1]. [4] Examiner insight: Refer to both the range of wavelengths and the transfer to the reaction centre.

10. (a) In photosystem II, in the thylakoid space [1]; products are electrons, protons and oxygen [1]. (b) Excited electrons from photosystem II pass along carriers [1]; the energy released moves protons out of the stroma and into the space inside the thylakoid [1]; photolysis adds more protons, so the space has a high proton concentration [1]; protons then move down their gradient into the stroma through ATP synthase, joining ADP to phosphate to give ATP [1]. (c) Two electrons from photosystem I [1] and a hydrogen ion from the stroma join NADP, forming reduced NADP [1]. [8] Examiner insight: Give the direction of proton movement at each step.

11. (a) 36 GP [1] (b) 36 TP are formed; 6 leave; so 30 TP regenerate RuBP [1] (c) 30 TP give 18 RuBP (5 TP to 3 RuBP: 30 ÷ 5 × 3) [1] (d) 6 TP ÷ 2 = 3 glucose [1] (e) ATP: 36 + 18 = 54 [1]; reduced NADP: 36 [1] (f) Five-sixths (30 ÷ 36) [1] [7] Examiner insight: Show a carbon check: 18 × 5 + 18 = 108 carbons = 36 × 3.

12. (a) RuBP: (35 − 100) ÷ 100 × 100 = −65% [1]; GP: (160 − 100) ÷ 100 × 100 = +60% [1] (b) In darkness neither ATP nor reduced NADP is produced [1]; GP cannot be converted to TP, so it accumulates [1]; RuBP is still used in fixation but not regenerated, so it falls [1]. (c) Without carbon dioxide, the Calvin cycle stops and reduced NADP is not used [1]; NADP is not regenerated, so electron carriers cannot pass on electrons [1]; photosystem II cannot emit electrons to the chain, so photolysis stops [1]. [8] Examiner insight: Link each effect step by step; “the cycle stops” alone does not explain photosystem II.

Where marks are usually lost

• Stating that oxygen comes from carbon dioxide.
• Taking Rf distances from the paper edge, or giving Rf units.
• Describing a spectrum without values.
• Mixing up absorption and action spectra.
• Naming a controlled variable but not saying how it is controlled.
• Dividing by the final value in a percentage change.
• Placing photolysis or carbon fixation in the wrong part of the chloroplast.

Next steps

• Revise with the revision notes and re-read the study guide.
• Open the IB DP Biology hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Topic C1.3 (Photosynthesis), statements C1.3.1 to C1.3.19; C1.3.9 to C1.3.19 are additional higher level.