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IB Biology C1.3: Photosynthesis worked solutions

Worked solutions to every practice question in C1.3: Photosynthesis of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]

Which statement about the oxygen released in photosynthesis is correct? (A) It comes from carbon dioxide. (B) It comes from the breakdown of glucose. (C) It is made in the light-independent reactions. (D) It comes from the photolysis of water.

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D. Oxygen gas is released when light splits water molecules (photolysis). Carbon dioxide supplies carbon, glucose is not broken down in photosynthesis, and the light-independent reactions do not release oxygen.

Final answer

Question A2

[1 mark]

Hydrogencarbonate indicator is red-orange at air level. A leaf in a sealed tube is in bright light for an hour and the indicator turns purple. What does this show? (A) The leaf stopped respiring. (B) The leaf respired faster than it photosynthesised. (C) The leaf photosynthesised faster than it respired. (D) The leaf stopped photosynthesising.

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C. Purple means the carbon dioxide concentration fell, so the leaf took in more carbon dioxide than it gave out: photosynthesis was faster than respiration. Respiration and photosynthesis both continue, so A and D claim too much, and B would turn the indicator yellow.

Final answer

Question A3

[1 mark]

A pigment looks red in white light. Which light does it absorb most strongly? (A) Red and orange (B) Red only (C) All wavelengths equally (D) Blue and green

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D. A red pigment reflects red light, so the light it absorbs is mainly blue and green. A is the light it reflects, B is wrong because it also absorbs other wavelengths, and C would make it look black.

Final answer

Question A4

[1 mark]

On a chromatogram the solvent front moved 9.0 cm from the origin and a pigment spot moved 5.4 cm from the origin. What is the R_f of the pigment? (A) 0.60 (B) 1.67 (C) 3.6 (D) 0.60 cm

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A. R_f=5.4\div9.0=0.60. R_f is a ratio with no unit, so D is wrong; B inverts the ratio and C is the difference in distances.

Final answer

Question A5

[1 mark]

At 25 ^{\circ}C and 0.04% CO_2 the rate of photosynthesis levels off at 6 units as light intensity rises. A second curve at 25 ^{\circ}C and 0.10% CO_2 levels off at 10 units. What limits the rate on the first plateau? (A) Temperature (B) Light intensity (C) Carbon dioxide concentration (D) Chlorophyll concentration

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C. On the plateau more light does not help, and the curve with more carbon dioxide reaches a higher rate at the same temperature, so carbon dioxide concentration limits the first plateau. Temperature is the same in both curves.

Final answer

Question A6

HL[1 mark]

Which event happens in photosystem II? (A) Water is split to replace electrons. (B) NADP is reduced. (C) Carbon dioxide is fixed. (D) RuBP is regenerated.

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A. Photolysis of water replaces the electrons that photosystem II loses. NADP is reduced at photosystem I, and carbon fixation and RuBP regeneration take place in the stroma.

Final answer

Question A7

HL[1 mark]

Which substances pass from the light-dependent reactions to the Calvin cycle? (A) Oxygen and carbon dioxide (B) ATP and reduced NADP (C) ADP and NADP (D) Glucose and RuBP

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B. ATP and reduced NADP are made in the light stage and used to reduce GP. ADP and NADP travel the other way, and oxygen leaves the chloroplast.

Final answer

Question A8

HL[1 mark]

A chloroplast fixes 9 molecules of carbon dioxide in the Calvin cycle. How many molecules of GP form? (A) 9 (B) 18 (C) 27 (D) 45

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B. Each carboxylation of RuBP gives two GP, so 9\times2=18. A counts the carbon dioxide and C and D multiply wrongly.

Final answer

Tutorial practice

Question B1

[5 marks]

The table shows the rate of oxygen release by two aquatic plants at three pH values. Table: pH 6, 7, 8; plant X 18, 15, 9; plant Y 12, 12, 11 (\mumol O_2 g^{-1} h^{-1}). (a) State the rate for plant X at pH 7, with its unit (\mumol O_2 g^{-1} h^{-1}). [1] (b) Calculate the percentage decrease in the rate of plant X between pH 6 and pH 8. [2] (c) Compare the effect of pH on the two plants. [2]

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(a) 15 with the unit \mumol g^{-1} h^{-1} (1). (b) (18-9)\div18\times100 (1) =50\% decrease (1). (c) A similarity: both fall as pH rises (1); a difference quoting values: X falls by 9 units and Y by only 1, so X is more sensitive to pH (1).

Final answer

(a) 15 \mumol g^{-1} h^{-1}. (b) 50%. (c) Both rates fall as pH rises; X falls much more than Y (X from 18 to 9, Y from 12 to 11); X is faster at pH 6 but slower than Y at pH 8, so Y is less affected.

Question B2

[5 marks]

A student measured the volume of oxygen released by a pondweed shoot in 15 minutes in light of one wavelength at a time. Table: wavelength 420, 480, 540, 600, 660, 720 nm; oxygen 45, 30, 9, 21, 42, 3 mm^3 in 15 min. (a) Calculate the rate of oxygen release at 420 nm and at 540 nm, with units. [2] (b) Calculate the rate at 540 nm as a percentage of the highest rate in the table. [1] (c) Explain why a graph of rate against wavelength is an action spectrum. [1] (d) Explain why the rate at 540 nm is low but not zero. [1]

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(a) 45\div15=3.0 (1) and 9\div15=0.6 mm^3 min^{-1} (1). (b) 0.6\div3.0\times100=20\% (1). (c) The y axis is the rate of oxygen release, which measures photosynthesis, so it is an action spectrum (1). (d) Chlorophyll reflects green light, so the rate in green is lower than in blue and red (1), but it is not zero because other pigments absorb some green light (1).

Final answer

(a) 3.0 and 0.6 mm^3 min^{-1}. (b) 20%. (c) It plots the rate of photosynthesis, not the light absorbed. (d) Chlorophyll reflects green light, so little green is absorbed; accessory pigments absorb a little, so the rate is low but not zero.

Question B3

[5 marks]

Discs cut from a leaf were floated in hydrogencarbonate solution under a lamp. Photosynthesis fills the air spaces of a disc with gas, so it rises. Students counted how many of 20 discs had risen after 10 minutes with the lamp at 10, 20 and 40 cm: 18, 11 and 4 discs. (a) Identify the independent and the dependent variable. [2] (b) State two variables that should be kept constant. [2] (c) Explain why the discs rise. [1]

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(a) Independent variable lamp distance or light intensity (1); dependent variable number of discs that have risen after a fixed 10 minutes (1). (b) Any two of: concentration of hydrogencarbonate, temperature, size and source of the discs, volume of solution (2). (c) Oxygen produced in photosynthesis collects in the air spaces of the disc, lowering its density so it rises; carbon dioxide does not (1).

Final answer

(a) Independent: lamp distance (light intensity); dependent: number of discs risen after 10 minutes. (b) Two of: same hydrogencarbonate concentration, temperature, disc size or leaf, time. (c) Oxygen from photosynthesis collects in the air spaces.

Question B4

[5 marks]

In a FACE experiment, open-air plots of young trees received extra carbon dioxide while untreated plots nearby acted as controls. The mean biomass gain over eight years was 2.0\pm0.3 kg C m^{-2} in the untreated plots and 2.6\pm0.3 kg C m^{-2} in the enriched plots. (a) Calculate the percentage increase in biomass gain caused by the extra carbon dioxide. [1] (b) State one advantage of FACE over an enclosed greenhouse. [1] (c) Explain how extra carbon dioxide could increase biomass. [2] (d) Suggest why a model could overpredict the biomass gain in a real forest. [1]

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(a) (2.6-2.0)\div2.0\times100=30\% (1). (b) FACE plots are in the open air in real conditions, unlike an enclosed chamber (1). (c) More carbon dioxide increases the rate of carbon fixation (1), so more carbohydrate is made and used for growth and biomass (1). (d) Another factor such as water, nutrients or light may limit the response in a real forest, so a model that assumes only CO_2 limits would overpredict (1).

Final answer

(a) 30%. (b) Conditions are those of a real forest. (c) More CO_2 gives faster fixation by Rubisco, so more carbohydrate for growth. (d) Light, water or nutrients may limit; models may not include them.

Question B5

[4 marks]

The rate of photosynthesis of a pond plant rises as the temperature goes from 5 to 30 ^{\circ}C, is highest at 30 ^{\circ}C and then falls to almost zero at 45 ^{\circ}C. Light and carbon dioxide are kept constant and plentiful. Explain the shape of the curve.

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Up to 30 ^{\circ}C temperature is the limiting factor (1); particles have more kinetic energy, so more collisions occur between substrate and the active sites of enzymes such as Rubisco (1). Above the optimum the enzymes begin to denature (1) and lose their active-site shape, so the rate falls steeply to almost zero (1).

Final answer

Rate rises because molecules move faster so enzyme-controlled steps such as carbon fixation speed up; temperature is limiting. The optimum is about 30 ^{\circ}C. Above it the enzymes, including Rubisco, denature, so the rate falls.

Question B6

[4 marks]

Pigments from a spinach leaf are to be separated on a silica gel plate. Describe how to prepare, run and read the chromatogram so that each pigment can be identified.

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Extract the pigments in an organic solvent such as propanone (1). Apply many small dry spots on a pencil origin line above the solvent level (1). Stand the plate in solvent in a closed container, remove it before the solvent reaches the top and mark the front (1). Measure from the origin line and calculate R_f for each spot; compare with colour and a reference (1).

Final answer

Extract with propanone; small dry spots on a pencil origin line above the solvent; solvent below the origin, plate in a closed container; mark the solvent front. Then R_f= distance of pigment \div distance of front, from the origin.

Question B7

HL[5 marks]

Light energy absorbed in a chloroplast is used to make ATP. Describe how the light energy is used to make ATP in the thylakoid membrane.

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Light is absorbed by pigments in photosystem II and excited electrons leave the reaction centre (1). The electrons pass along carriers in the thylakoid membrane (1). The energy released pumps protons from the stroma into the thylakoid space (1). Photolysis replaces the lost electrons and adds protons to the space (1). Protons flow back through ATP synthase down the gradient and ADP is phosphorylated to ATP (1).

Final answer

Light excites electrons in photosystem II; they pass along carriers in the thylakoid membrane, releasing energy that pumps protons into the thylakoid space; photolysis replaces the electrons and adds protons; protons flow back through ATP synthase, joining ADP and Pi (chemiosmosis).

Question B8

HL[5 marks]

Carbon dioxide from the air is converted to carbohydrate in the stroma, which contains a great deal of Rubisco. Explain the conversion and the quantity of Rubisco.

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Rubisco catalyses carboxylation: carbon dioxide joins ribulose bisphosphate and the product splits into two GP (1). GP is reduced to triose phosphate (1) using reduced NADP and ATP from the light-dependent reactions (1). Most triose phosphate is used to regenerate RuBP with ATP, and some leaves to make carbohydrate (1). Rubisco works slowly and is inefficient at low CO_2, so a large amount is needed (1).

Final answer

Rubisco joins CO_2 to RuBP, giving two GP; GP is reduced to TP using ATP and reduced NADP; most TP regenerates RuBP using ATP, and one-sixth leaves as carbohydrate. Rubisco is slow and poor at low CO_2 so much is needed.

Question B9

HL[4 marks]

A chloroplast and a mitochondrion both make ATP at a membrane. Compare ATP production at the thylakoid membrane with that at the inner mitochondrial membrane, giving two similarities and two differences.

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Similarity: both use an electron transport chain to pump protons across a membrane (1) and both use ATP synthase to make ATP by chemiosmosis (1). Difference, any two for one mark each: electrons from water or from reduced NAD; final acceptor NADP or oxygen; energy to excite electrons from light or from food (max 2).

Final answer

Similar: both pump protons across a membrane by electron transport and make ATP in ATP synthase by chemiosmosis. Different: electrons come from water (chloroplast) or reduced NAD (mitochondrion); final acceptor NADP or oxygen; light energy or food energy.

Question B10

HL[3 marks]

Photosystems are found in thylakoid membranes. Outline the structure and function of a photosystem.

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A photosystem is a protein complex in the thylakoid membrane holding many chlorophyll and accessory pigment molecules (1). The pigments absorb light of a range of wavelengths and pass the energy to a reaction centre chlorophyll (1), which emits an excited electron (1).

Final answer

A protein complex in a membrane with many chlorophyll and accessory pigments around a reaction centre; pigments absorb different wavelengths and pass energy to the reaction centre, which emits an excited electron.

Advanced practice

Question C1

[7 marks]

Net carbon dioxide uptake was measured for leaves of a sun plant S and a shade plant T at four light intensities. Negative values mean net release. Assume that respiration is the same at every light intensity. Table: light 0, 50, 100, 200; plant S -1.5, 2.5, 6.0, 8.5; plant T -3.0, -0.5, 1.0, 1.5 (\mumol CO_2 m^{-2} s^{-1}). (a) State between which two light intensities plant T reaches its balance point. [1] (b) Calculate the gross photosynthesis of each plant at 200 units, with the units (\mumol CO_2 m^{-2} s^{-1}). [2] (c) Compare the response of the two plants to increasing light. [2] (d) Explain what limits plant T between 100 and 200 units, and suggest a difference between the plants. [2]

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(a) For T the sign changes between 50 and 100 units (1). (b) The value at zero light is respiration alone, 3.0, so gross photosynthesis for T at 200 is 1.5+3.0=4.5; for S at 200 it is 8.5+1.5=10.0 (1 for each, 2). (c) Both rise as light increases (1); S rises much more, from -1.5 to 8.5 (a change of 10.0) while T rises from -3.0 to 1.5 (a change of 4.5) (1). (d) T's curve is flattening between 100 and 200 so another factor limits it, whereas S is still rising and light is still limiting (1); credit a link to the different chlorophyll or enzyme content (1).

Final answer

(a) Between 50 and 100 units. (b) S 10.0, T 4.5 \mumol CO_2 m^{-2} s^{-1}. (c) Both rise; S rises by 10.0 and T by only 4.5. (d) T is flattening, so another factor limits it; S is still limited by light.

Question C2

HL[7 marks]

A mutant alga cannot make functional photosystem II but has working photosystem I, carriers and ATP synthase. It is kept in light with plenty of carbon dioxide. (a) Predict whether the mutant releases oxygen, and explain. [1] (b) Describe how the mutant can still make ATP. [2] (c) Explain why the mutant cannot make carbohydrate. [2] (d) Suggest why a normal alga also benefits from cyclic electron flow. [2]

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(a) No oxygen is released because photolysis occurs at photosystem II (1). (b) Cyclic electron flow: electrons from photosystem I return to the carriers (1), pumping protons so ATP synthase can make ATP (1). (c) No electrons arrive from water, so NADP is not reduced (1); GP cannot be reduced to triose phosphate without reduced NADP, so carbohydrate is not made (1). (d) The Calvin cycle needs three ATP for every two reduced NADP, so cyclic flow supplies extra ATP (1) without using photosystem II or making more reduced NADP (1).

Final answer

(a) No oxygen: no photolysis at PS II. (b) Cyclic flow at PS I; electrons return to carriers; protons pumped, ATP made. (c) No electrons from water, so NADP is not reduced; GP cannot be reduced, so no carbohydrate. (d) Wild type needs extra ATP because the cycle uses 3 ATP per 2 reduced NADP.

Question C3

HL[7 marks]

A lit culture of algae was supplied with carbon dioxide. At time 0 the carbon dioxide was removed. The table shows the relative amounts of GP and RuBP and the rate of oxygen release. Table: minutes 0, 2, 4; relative GP 100, 40, 10; relative RuBP 100, 180, 230; O_2 release 100, 55, 20 percent. (a) Describe the changes in GP and RuBP. [2] (b) Explain the changes. [2] (c) Explain the fall in oxygen release. [3]

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(a) GP falls from 100 to 10 (1) while RuBP rises from 100 to 230 (1). (b) Without carbon dioxide RuBP is not carboxylated, so no new GP forms and the existing GP is used up (1); RuBP is still regenerated from triose phosphate but is not used, so it accumulates (1). (c) ATP and reduced NADP are no longer used, so NADP is not returned (1); the carriers become full of electrons and electron flow from photosystem II slows (1), so photolysis slows and less oxygen is released (1).

Final answer

(a) GP falls 100 to 10; RuBP rises 100 to 230. (b) RuBP is no longer carboxylated so GP is not made, while TP still regenerates RuBP. (c) Without GP reduction NADP is not regenerated, so carriers fill and PSII stops; photolysis and oxygen release fall.

Question C4

HL[7 marks]

Algae in light were supplied with ^{14}CO_2 for a short time. Each molecule of fixed CO_2 supplies one labelled carbon to one of the two GP molecules formed from each RuBP. Six ^{14}CO_2 molecules were fixed. (a) State how many of the GP molecules formed are labelled, and how many GP molecules form in total. [1] (b) Deduce how many triose phosphate molecules are labelled, with a reason. [2] (c) Explain why RuBP becomes labelled. [2] (d) Show by counting carbons how many TP leave and how many RuBP are regenerated. [2]

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(a) Six GP molecules are labelled, out of 12 formed (1). (b) Reduction keeps the carbon skeleton, so 6 of the 12 TP are labelled (1) because each GP gives one TP (1). (c) Some labelled TP is used to regenerate RuBP (1), so RuBP also becomes labelled (1). (d) Two TP leave, 2\times3=6 carbons, enough for one glucose (1); the other 10 TP hold 30 carbons, which rebuild 30\div5=6 RuBP (1).

Final answer

(a) 6 labelled GP out of 12. (b) 6 labelled TP; labelled carbon passes to TP in reduction. (c) The labelled TP regenerates RuBP, so RuBP carries label. (d) 2 TP leave: 6 C; 10 TP (30 C) rebuild 6 RuBP (30 C).

Question C5

[7 marks]

Discs of a seaweed were placed in seawater with a low or a high concentration of hydrogencarbonate at four light intensities. The table shows the rate of oxygen release. Table: light 10, 30, 60, 100; low hydrogencarbonate 2, 5, 6, 6; high 2, 5, 9, 11 (\mumol O_2 g^{-1} h^{-1}). (a) State the limiting factor for the low-hydrogencarbonate series at 10 and at 100 units, with a reason for each. [2] (b) State what limits the rate of each series at 60 units. [2] (c) Calculate the percentage increase in rate caused by the extra hydrogencarbonate at 100 units. [2] (d) State one variable that should be kept constant in this experiment. [1]

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(a) At 10 units light limits both series (1); at 100 units with low hydrogencarbonate the rate has levelled at 6, so carbon supply, not light, limits (1). (b) At 60 units the low series has levelled at 6 while the high series is higher at 9, so hydrogencarbonate limits the low series (1); the high series is still rising, so light limits it (1). (c) (11-6)\div6\times100=83\% (2; 1 for the working, 1 for the answer). (d) Keep the temperature constant with a water bath, because the lamp warms the water (1).

Final answer

(a) Light at 10 units; at 100 units the low series has levelled, so hydrogencarbonate (carbon supply) limits. (b) Low: hydrogencarbonate; high: light. (c) 83%. (d) Temperature.

Connected questions

Question X1

Connected questions · HL[6 marks]

C1.3 + C1.2. A potted plant in constant dim light shows zero net exchange of carbon dioxide with the air. (a) State what zero net exchange shows about the two processes. [1] (b) State two ways in which ATP is made similarly in a chloroplast and a mitochondrion. [2] (c) Explain why removing oxygen from the air would stop ATP production in the mitochondria but not stop the light-dependent reactions at once. [3]

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(a) The rate of photosynthesis equals the rate of respiration, so this is the balance point (1). (b) Two of: electron carriers pass electrons along a chain; protons are pumped into a space; the proton gradient drives ATP synthase; ATP is made from ADP and phosphate (2). (c) Mitochondria need oxygen as the final electron acceptor, so ATP production there falls without oxygen (1); the light-dependent reactions use light and water and do not use oxygen, so they continue (1); their ATP comes from a proton gradient made by light-excited electrons, which needs no oxygen (1).

Final answer

(a) Photosynthesis equals respiration. (b) Both pump protons across a membrane and use ATP synthase; electrons are passed along carriers. (c) Chloroplast light stage needs light and no oxygen; mitochondrion needs oxygen as the final acceptor, so ATP production there would stop.

Question X2

Connected questions[6 marks]

C1.3 + C1.1. A greenhouse crop receives constant light. The rate of carbon fixation rises between 5 and 25 ^{\circ}C and falls above 35 ^{\circ}C. (a) Explain why the rate rises with temperature. [2] (b) Explain why the rate falls above 35 ^{\circ}C. [2] (c) Explain why raising the CO_2 concentration raises the rate at 25 ^{\circ}C but has little effect at 5 ^{\circ}C. [2]

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(a) Molecules have more kinetic energy, so collisions between Rubisco and its substrates are more frequent and forceful (1) and the enzyme-controlled steps speed up (1). (b) Above the optimum the enzymes begin to denature (1) and the active site loses its shape so the substrate no longer fits (1). (c) At 5 ^{\circ}C enzyme activity is the limiting factor, so extra carbon dioxide cannot be used faster (1); at 25 ^{\circ}C the enzymes work faster and the supply of carbon dioxide limits, so more carbon dioxide raises the rate (1).

Final answer

(a) More kinetic energy so more enzyme-substrate collisions up to the optimum. (b) Above the optimum enzymes denature, losing active-site shape. (c) At 5 C enzymes are slow, so temperature limits, and more CO_2 cannot raise the rate; at 25 C CO_2 limits.

Question X3

Connected questions[6 marks]

C1.3 + C4.2. In a meadow the plants fix 12 000 kJ m^{-2} yr^{-1} of light energy as chemical energy in carbon compounds. They use 4 500 kJ m^{-2} yr^{-1} of this in respiration. (a) Calculate the energy remaining for growth, with units. [1] (b) Calculate the percentage of the energy fixed that is used in respiration. [1] (c) State two fates of the energy that remains. [2] (d) Suggest how rising atmospheric CO_2 could affect carbon storage in the meadow. [2]

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(a) 12\,000-4\,500=7\,500 kJ m^{-2} yr^{-1} (1). (b) 4\,500\div12\,000\times100=37.5\% (1). (c) Two of: stored in the biomass of the plants; transferred to herbivores that eat them; passed to decomposers when plants die (2). (d) Extra carbon dioxide can raise the rate of carbon fixation (1), so more carbon is stored in biomass, provided other factors do not limit growth (1).

Final answer

(a) 7500 kJ. (b) 37.5%. (c) Growth stored in plant biomass; eaten by herbivores or passes to decomposers. (d) More CO_2 means more fixation, so more carbon stored in biomass, if nothing else limits.

Question X4

Connected questions · HL[6 marks]

C1.3 + B2.2. Thylakoid membranes in a chloroplast are folded into stacks that enclose a very small thylakoid space. A chemical makes the thylakoid membrane leak protons freely. (a) Explain why a small thylakoid space helps ATP production. [2] (b) Predict and explain the effect of the leaky membrane on ATP production. [2] (c) Predict the effect on the Calvin cycle. [2]

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(a) A small volume means that pumping a few protons produces a large rise in concentration, so a steep gradient forms (1) with a high concentration of protons inside (1). (b) The proton gradient is lost (1), so protons do not flow through ATP synthase and little ATP is made (1). (c) With little ATP, GP is not reduced to triose phosphate or RuBP regenerated (1), so carbohydrate synthesis falls (1).

Final answer

(a) A small space means few protons raise the concentration steeply. (b) The gradient collapses, so ATP synthase makes little ATP. (c) Without ATP the Calvin cycle cannot reduce GP, so carbohydrate falls; electron flow and oxygen release continue.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Photosynthesis from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book