3.1- Photosynthesis

Question 1

Q1.Scientists studied the rate of carbon dioxide uptake by grape plant leaves. Grape leaves
have stomata on the lower surface but no stomata on the upper surface.
The scientists recorded the carbon dioxide uptake by grape leaves with three different
treatments:
Treatment 1 − No air-sealing grease was applied to either surface of the leaf.
Treatment 2 − The lower surface of the leaf was covered in air-sealing grease that
prevents gas exchange.
Treatment 3 − Both the lower surface and the upper surface of the leaf were covered in
air–sealing grease that prevents gas exchange.
The scientists measured the rate of carbon dioxide uptake by each leaf for 60 minutes in
light and then for 20 minutes in the dark.
The scientists’ results are shown in the diagram below.
(a) Suggest the purpose of each of the three leaf treatments (3)

Answer 1

1. (No grease)
 means stomata are open
 OR
 allows normal CO2 uptake;
Allow ‘gas exchange’ for CO2 uptake.
‘As a control’ is insufficient on its own.
2. (Grease on lower surface)
seals stomata
OR
stops CO2 uptake through
stomata
OR
to find CO2 uptake through
stomata
OR
shows CO2 uptake through cuticle / upper surface;
3. (Grease on both surfaces) shows sealing is effective
OR
stops all CO2 uptake.

Question 2

In the light-dependent reaction of photosynthesis, light energy generates ATP. Describe how. (5)

Answer 2

Light energy raises energy level of electrons in chlorophyll;

Electrons leave chlorophyll and pass down electron transfer chain;

The passage involves a series of redox reactions;

Energy is released by electrons at each carrier;

This energy is used to generate ATP from ADP and phosphate

Question 3

In the light-independent reaction of photosynthesis, the carbon in carbon dioxide becomes carbon in triose phosphate. Describe how. (5)

Answer 3

Carbon dioxide combines with ribulose bisphosphate;

To produce two molecules of glycerate 3-phosphate;

Which is reduced to triose phosphate;

Which requires reduced NADP;

and energy from ATP

TP converted into gluscose

Question 4

The diagram above shows the LDR of photosynthesis
a) what does object a represent? (1)
b)describe the process of B and its purpose (3)
C) Explain how reactant D is made into reduced NADP ? (2)

CGP- pg 117

Answer 4

a) Photosystem II

b)Photolysisn/light energy
split water into 2 hydrogen ions and oxygen
the electrons from the water replace the electrons lost from the chlorophyll

c)excited electrons are transferred to reactant NADP from photosystem I/object C

along with a proton/ H+ ion from the stroma

Question 5

Q2.A student investigated the effect of different wavelengths of light on the rate of
photosynthesis. She used the apparatus shown in Figure 1.

(a) What measurements should the student have taken to determine the rate of
photosynthesis?
(1)

(b) Other than temperature and pH, give two factors which should be kept constant
during this investigation.

Answer 5

(a) Oxygen production / concentration and time.
Accept: oxygen volume / concentration
Reject: oxygen uptake
Neutral: reference to carbon dioxide uptake

b)1. Intensity of light;
Accept: distance from light
2. Amount / number / mass / species of algae / photosynthesising cells;
3. Carbon dioxide (concentration / partial pressure);
4. Time

Question 6

Q2.A student investigated the effect of different wavelengths of light on the rate of
photosynthesis. She used the apparatus shown in Figure 1.

(c) The student did not use a buffer to maintain the pH of the solution.
Explain what would happen to the pH of the solution during this investigation. (2)

(d) Figure 2 shows the student’s results.
Figure 2
Suggest and explain why the rate of photosynthesis was low between 525 nm and
575 nm wavelengths of light. (2)

Answer 6

(c) 1. (pH) increases;
Neutral: becomes more alkaline / less acidic
2. As (more) carbon dioxide removed (for photosynthesis).

1. Less absorption / (more) reflection (of these wavelengths of light);
   Reject: no absorption or cannot absorb unless in context of
   green light.
   Note: no green light absorbed or green light reflected = 2
   marks.
2. (Light required) for light dependent (reaction) / photolysis
   Accept: for excitation / removal of electrons (from
   chlorophyll)
3. (Represents) green light / colour of chlorophyll.

Question 7

Q3.(a) On islands in the Caribbean, there are almost 150 species of lizards belonging to the
genus Anolis. Scientists believe that these species evolved from two species found
on mainland USA. Explain how the Caribbean species could have evolved.
(6)

Answer 7

1. Geographic(al) isolation;
2. Separate gene pools / no interbreeding / gene flow (between
   populations);
   Accept: reproductive isolation
   This mark should only be awarded in context of during the
   process of speciation. Do not credit if context is after
   speciation has occurred.
3. Variation due to mutation;
4. Different selection pressures / different abiotic / biotic conditions /
   environments / habitats;
   Neutral: different conditions / climates if not qualified
   Accept: named abiotic / biotic conditions
5. Different(ial) reproductive success / selected organisms (survive and)
   reproduce;
   Accept: pass on alleles / genes to next generation as
   equivalent to reproduce
6. Leads to change / increase in allele frequency.
   Accept: increase in proportion / percentage as equivalent to
   frequency

Question 8

(b) Anolis sagrei is a species of lizard that is found on some of the smallest Caribbean
islands. Describe how you could use the mark-release-recapture method to estimate
the number of Anolis sagrei on one of these islands.
(4)

Answer 8

1. Capture / collect sample, mark and release;
2. Method of marking does not harm lizard / make it more visible to
   predators;
3. Leave sufficient time for lizards to (randomly) distribute (on island)
   before collecting a second sample;
4. (Population =) number in first sample × number in second sample
   divided by number of marked lizards in second sample / number
   recaptured.

Question 9

(c) Large areas of tropical forest are still found on some Caribbean islands. The
concentration of carbon dioxide in the air of these forests changes over a period of
24 hours and at different heights above ground.
Use your knowledge of photosynthesis and respiration to describe and explain how
the concentration of carbon dioxide in the air changes:
• over a period of 24 hours
• at different heights above ground. (5)

Answer 9

1. High concentration of / increase in carbon dioxide linked with respiration
   at night / in darkness;
2. No photosynthesis in dark / night / photosynthesis only in light / day;
   Neutral: less photosynthesis
3. In light net uptake of carbon dioxide / use more carbon dioxide than
   produced / (rate of) photosynthesis greater than rate of respiration;
4. Decrease in carbon dioxide concentration with height;
   More carbon dioxide absorbed higher up
   Accept: less carbon dioxide higher up / more carbon dioxide
   lower down
5. (At ground level)
   less photosynthesis / less photosynthesising tissue / more respiration /
   more micro-organisms / micro-organisms produce carbon dioxide.
   Neutral: less leaves unqualified or reference to animals

Question 10

Q4.Chloroplasts contain chlorophyll a and chlorophyll b. Scientists found tobacco plants with a
mutation that caused them to make more chlorophyll b than normal tobacco plants. They
investigated the effect of this mutation on the rate of photosynthesis.

The scientists carried out the following investigation.
• They grew normal and mutant tobacco plants. They grew some of each in low light
intensity and grew others in high light intensity.
• They isolated samples of chloroplasts from mature plants of both types.
• Finally, they measured oxygen production by the chloroplasts they had isolated from
the plants.
The figure below shows the scientists’ results.

(a) Explain why the scientists measured the rate of production of oxygen in this
investigation. (2)

(b) Calculate the difference in the oxygen produced by the chloroplasts from mutant
plants grown in low and high light intensities at a light intensity of 500 μmol photons
m–2 s–1
.
Show your working. (2)

Answer 10

1. Oxygen produced in light-dependent reaction;
2. The faster (oxygen) is produced, the faster the light-dependent reaction

(b) 35–36 μmol Oxygen per mg chlorophyll.
Correct difference at 500 μmol photons m–2 s–1 or incorrect
difference but division by 4 shown = 1 mark.

Question 11

(c) The scientists suggested that mutant plants producing more chlorophyll b would
grow faster than normal plants in all light intensities.
Explain how these data support this suggestion.
…………………………………………………………………………………. (4)

Answer 11

(c) At all light intensities, chloroplasts from mutant plants:
1. Have faster production of ATP and reduced NADP;
2. (So) have faster / more light-independent reaction;
3. (So) produce more sugars that can be used in respiration;
4. (So) have more energy for growth;
5. Have faster / more synthesis of new organic materials.
Accept converse points if clear answer relates to non-mutant
plants

Question 12

Q5.Farmland previously used for growing crops was left for 30 years and developed into
woodland. During this period, ecologists recorded an increase in the diversity of birds in
the area.
(a) Name the process that resulted in the development of woodland from farmland.

b) Explain the increase in the diversity of birds as the woodland developed.

Answer 12

5.(a) Succession;
Ignore any word in front of succession e.g. secondary /
ecological succession.
Neutral ‘forestation’.

b) 1. Greater variety / diversity of plants / insects / more plant / insect species;
Neutral: more plants.
2. More food sources / more varieties of food;
Neutral: more food / more / greater food source (singular).
3. Greater variety / more habitats / niches;
Accept: more nesting sites.
Q Neutral: more homes / shelters.

Question 13

(c) The ecologists also investigated photosynthesis in two species of plant found in the
woodland. One of the species was adapted to growing in bright sunlight (sun plant)
and the other was adapted to growing in the shade (shade plant). The ecologists’
results are shown in the figure below.
(i) Give two factors which could be limiting the rate of photosynthesis in the sun
plant between points A and B on the figure

Answer 13

(i) Temperature and carbon dioxide;

Neutral: water, chlorophyll.

Question 14

(ii) Explain why CO2 uptake is a measure of net productivity.

Answer 14

(ii) Shows (gross) photosynthesis / productivity minus respiration / more
carbon dioxide used in photosynthesis than produced in respiration;
Correct answers are often shown as: net productivity =
(gross) photosynthesis – (minus) respiration.

Question 15

(iii) Use the information in the figure to explain how the shade plant is better
adapted than the sun plant to growing at low light intensities.

Answer 15

(iii) 1. (Shade plant) has lower (rate of) respiration / respiratory losses /
less CO2 released at 0 light intensity / in dark;
Accept use of figures.
Accept: lower compensation point.
2. Greater (net) productivity / less sugars / glucose used / more
sugars / glucose available;
Neutral: any references to rate of photosynthesis.

Question 16

0 1 Heat stress is a condition that often occurs in plants exposed to high temperatures for
a prolonged period of time. Heat stress is a major factor in limiting the rate of
photosynthesis.
0 1 . 1 Heat stress decreases the light-dependent reaction of photosynthesis.
Explain why this leads to a decrease in the light-independent reaction.
[2 marks]

Answer 16

1. (Less/no) ATP;
2. (Less/no) reduced NADP;

Approximately 41% of students obtained both marks by referring to the reduction in ATP
and reduced NADP. Students who gained a single mark usually did so by stating that there
was less ATP rather than less reduced NADP. Many of these students incorrectly stated
that less NADP or less reduced NAD was produced. Some students suggested that heat
stress caused stomata to close and this limited carbon dioxide uptake and photosynthesis.
Many students used additional pages for this question due to initially describing
denaturation of rubisco or ATP synthase before realising that they needed to name specific
products of the light-dependent reaction. It was disappointing to note that nearly a third of
students scored zero.

Question 17

0 1 Heat stress is a condition that often occurs in plants exposed to high temperatures for
a prolonged period of time. Heat stress is a major factor in limiting the rate of
photosynthesis.

1 . 2 Another effect of heat stress is a decrease in the activity of the enzyme rubisco. A
decrease in the activity of an enzyme means that the rate of the reaction it catalyses
becomes slower.
A decrease in the activity of the enzyme rubisco would limit the rate of photosynthesis.
Explain why [2 marks]

Answer 17

1. (Less/no) carbon dioxide (reacts) with RuBP;
2. (Less/no) GP;

failed to mention both RuBP and
carbon dioxide, or simply referred to a six-carbon product with no mention of GP being
formed

A minority of students incorrectly referred to GP as glucose phosphate.

Question 18

1 . 3 Where precisely is rubisco found in a cell?

[1 mark]

Answer 18

1. Stroma (of/in chloroplast);

01.3 Almost two-thirds of students correctly named the stroma as the location of rubisco. The
thylakoids, crista and matrix were common incorrect responses

Question 19

Scientists investigated the effect of temperature on the activity of two enzymes box
isolated from the leaf cells of cotton plants.
• Rubisco
• Rubisco activase – an enzyme that activates rubisco
Figure 1 and Figure 2 show their results.
Figure 1 Figure 2

0 1 . 4 The scientists concluded that heat stress reduces the activity of rubisco in plant
leaves by affecting rubisco activase.
Use all the information to evaluate their conclusion.
[4 marks]

Answer 19

1. Rubisco activity increases with temperature
   OR
   Rubisco optimum temperature is above
   (rubisco activase);
2. (Rubisco) activase activity decreases at
   high temperatures (allow any temperature
   above 25 ºC.)
   OR
   (Rubisco) activase optimum (allow in range)
   25 to 30 ºC.;
3. (Results/graphs suggest) activase
   cannot/does not affect activity of rubisco;
4. (Results are) only for cotton;
5. (Results are) for isolated enzymes;
6. No stats test

01.4 The majority of students (56%) obtained two marks, invariably for describing the effects of
an increase in temperature on the activity of rubisco and rubisco activas
e.

use all the information successfully to evaluate the scientists’
conclusion.

Only 25% of students obtained more than two of the four marks available.
Relatively few students clearly stated that the results indicate that rubisco activase does not
activate rubisco. Similarly, very few students stated that these results were only for cotton
plants and were for isolated enzymes. Equally surprising was the scarcity of responses
which referred to the lack of a statistical test. Consequently, this question did not
discriminate as effectively as had been expected.

Question 20

7 box . 1 In photosynthesis, which chemicals are needed for the light-dependent reaction?
Tick () one box.
[1 mark]
Reduced NADP, ADP, Pi, water and oxygen.

NADP, ATP and water.

Reduced NADP, ATP, water and carbon dioxide.

NADP, ADP, Pi and water.

Answer 20

 NADP, ADP, Pi and water;

07.1 Almost 70% of students correctly identified the chemicals needed for the light-dependent
reaction.

Question 21

0 7 . 2 Describe what happens during photoionisation in the light-dependent reaction.
[2 marks]

Answer 21

1. Chlorophyll absorbs light
   OR
   Light excites/moves electrons in chlorophyll;
2. Electron/s are lost
   OR
   (Chlorophyll) becomes positively charged;

07.2 Photoionisation in the light-dependent reaction was clearly described by many students,
with 44% obtaining both marks and approximately 77% obtaining at least one mark. Some
students limited their description to photolysis and gained no marks. Many students
referred to chlorophyll absorbing light and/or light exciting the electrons in chlorophyll.
Some answers referring only to chloroplasts or photosystems did not obtain this mark point.
Many students referred to electrons being lost or being passed to the electron transport
chain. Few students referred to chlorophyll molecules becoming positively charged.

Question 22

A student obtained a solution of pigments from the leaves of a plant. Then the
student used paper chromatography to separate the pigments.
Figure 5 shows the chromatogram produced.
Figure 5
0 7 . 3 Explain why the student marked the origin using a pencil rather than using ink.
[1 mark]

Answer 22

Ink and (leaf) pigments would mix
OR
(With ink) origin/line in different position
OR
(With pencil) origin/line in same position
OR
(With pencil) origin/line still visible;

07.3 This question was not well answered and was a poor discriminator. Slightly more than 20%
of students obtained the mark. Only the minority of students could clearly explain that
pencil was used to mark the origin so that the origin/line was still visible after running the
chromatogram. The problem of ink dissolving in the solvent was appreciated by many but
the consequences of this were not explained; the mixing of the ink and leaf pigments was
not appreciated by most students. Answers often only referred to the ink
running/smudging/dissolving in the solvent and affecting results. It was not always clear
whether students were writing about pencil or ink

Question 23

0 7 . 4 Describe the method the student used to separate the pigments after the
solution of pigments had been applied to the origin.
[2 marks]

Answer 23

1. Level of solvent below origin/line;
2. Remove/stop before (solvent) reaches top/end;

only mentioned placing the
chromatogram in a solvent, adding a lid and running the chromatogram for a set time or
until the pigments had separated. More students obtained a mark for the level of the
solvent being below the origin than for marking the solvent front or for removing the
chromatogram before it reached the top/end. Some students explained the principle of
chromatography rather than providing details on the method. A range of inappropriate
solvents were referred to including water, glucose and hydrochloric acid.

Question 24

0 7 . 6 The pigments in leaves are different colours. Suggest and explain the advantage of
having different coloured pigments in leaves.
[1 mark]

Answer 24

(Absorb) different/more wavelengths (of light) for
photosynthesis;

07.6 Approximately 46% of students obtained this mark, explaining that different
wavelengths/frequencies of light could be absorbed for photosynthesis. Students failing to
gain credit often omitted to mention photosynthesis or did not refer to wavelength or
frequency of light. A few students suggested the different colours of pigments enabled
them to be identified during chromatography.

Question 25

0 4 A student isolated chloroplasts from spinach leaves into a solution to form a
chloroplast suspension. He used the chloroplast suspension and DCPIP solution
to investigate the light-dependent reaction of photosynthesis. DCPIP solution is
blue when oxidised and colourless when reduced.
The student set up three test tubes as follows:
• Tube 1 – 1 cm3 of solution without chloroplasts and 9 cm3 of DCPIP solution in
light.
• Tube 2 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in
darkness.
• Tube 3 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in light.
The student recorded the colour of the DCPIP in each of the tubes at the start and
after the tubes had been left at 20 °C for 30 minutes.
His results are shown in Table 1.
Table 1
Tube
Colour of DCPIP in tube
At start After 30 minutes
1 blue blue
2 blue blue
3 blue colourless
0 4 . 1 The solution that the student used to produce the chloroplast suspension had the
same water potential as the chloroplasts.
Explain why it was important that these water potentials were the same.
[2 marks]

Answer 25

1. Osmosis does not occur;
2. Chloroplast/organelle does not
   burst/lyse/shrivel/shrink;

1. Accept: osmosis would occur if
water potentials were not the
same.
1 and 2, Accept: correct
reference to osmotic lysis for 2
marks.
2. Accept: chloroplast would
burst/lyse/shrivel/shrink if water
potentials were not the same.
2. Reject: ‘cell bursts/shrivels’
2. Ignore: damage to
chloroplasts on its own is not
enough for a mark.
2. Reject: becomes turgid/flaccid.

04.1 This was generally well answered, with over 80% of students obtaining at least one mark.
Most students understood that using a solution of the same water potential would prevent
osmosis, and then described the effects on chloroplasts if the water potentials were not the
same. Common errors included discussing the effect of osmosis on ‘the cell’ rather than on
chloroplasts and the use of terms such as ‘plasmolysed’ and ‘turgid’. The need for water in
photolysis or the light-dependent reaction was also referred to in weaker responses. Some
students suggested that the movement of water had to be prevented to avoid changing the
colour of the DCPIP.

Question 26

0 4 A student isolated chloroplasts from spinach leaves into a solution to form a
chloroplast suspension. He used the chloroplast suspension and DCPIP solution
to investigate the light-dependent reaction of photosynthesis. DCPIP solution is
blue when oxidised and colourless when reduced.
The student set up three test tubes as follows:
• Tube 1 – 1 cm3 of solution without chloroplasts and 9 cm3 of DCPIP solution in
light.
• Tube 2 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in
darkness.
• Tube 3 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in light.
The student recorded the colour of the DCPIP in each of the tubes at the start and
after the tubes had been left at 20 °C for 30 minutes.
His results are shown in Table 1.
Table 1
Tube
Colour of DCPIP in tube
At start After 30 minutes
1 blue blue
2 blue blue
3 blue colourless
0 4 . 1 The solution that the student used to produce the chloroplast suspension had the
same water potential as the chloroplasts.

4 . 2 Explain why the student set up Tube 1. (BLUE TO BLUE)
[2 marks]

Answer 26

1. To show light does not affect DCPIP;
2. To show chloroplasts are required;

04.2 This question proved to be more demanding with less than 10% of students obtaining both
marks. Many answers simply referred to Tube 1 as being a control, or that it was set up to
allow comparison with other tubes. Some students suggested that it was the DCPIP
causing the colour change. However, almost 60% of students gained at least one mark,
usually for describing that Tube 1 showed that chloroplasts were needed to cause the
colour change. The idea that Tube 1 shows that light does not affect DCPIP was less often
mentioned.

Question 27

0 4 A student isolated chloroplasts from spinach leaves into a solution to form a
chloroplast suspension. He used the chloroplast suspension and DCPIP solution
to investigate the light-dependent reaction of photosynthesis. DCPIP solution is
blue when oxidised and colourless when reduced.
The student set up three test tubes as follows:
• Tube 1 – 1 cm3 of solution without chloroplasts and 9 cm3 of DCPIP solution in
light.
• Tube 2 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in
darkness.
• Tube 3 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in light.
The student recorded the colour of the DCPIP in each of the tubes at the start and
after the tubes had been left at 20 °C for 30 minutes.
His results are shown in Table 1.
Table 1
Tube
Colour of DCPIP in tube
At start After 30 minutes
1 blue blue
2 blue blue
3 blue colourless
0 4 . 1 The solution that the student used to produce the chloroplast suspension had the
same water potential as the chloroplasts.

0 4 . 3 Explain the results in Tube 3.
[2 marks]

Answer 27

1. Reduction of DCPIP by electrons;
2. (From) chlorophyll/light dependent
   reaction;

04.3 Over 80% of students obtained at least one mark for this question, usually for outlining the
role of chlorophyll or the light-dependent reaction in producing the results in Tube 3. Better
students did refer to the reduction of DCPIP by electrons. However, common
misconceptions suggested that DCPIP was reduced by protons or by reduced NADP (or
reduced NAD). Weaker responses referred to the oxidation of DCPIP by oxygen or the
release of ions from chlorophyll.

Question 28

0 4 A student isolated chloroplasts from spinach leaves into a solution to form a
chloroplast suspension. He used the chloroplast suspension and DCPIP solution
to investigate the light-dependent reaction of photosynthesis. DCPIP solution is
blue when oxidised and colourless when reduced.
The student set up three test tubes as follows:
• Tube 1 – 1 cm3 of solution without chloroplasts and 9 cm3 of DCPIP solution in
light.
• Tube 2 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in
darkness.
• Tube 3 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in light.
The student recorded the colour of the DCPIP in each of the tubes at the start and
after the tubes had been left at 20 °C for 30 minutes.
His results are shown in Table 1.
Table 1
Tube
Colour of DCPIP in tube
At start After 30 minutes
1 blue blue
2 blue blue
3 blue colourless
0 4 . 1 The solution that the student used to produce the chloroplast suspension had the
same water potential as the chloroplasts.

0 4 . 4 The student evaluated the effectiveness of different chemicals as weed-killers by
assessing their ability to prevent the decolourisation of DCPIP in chloroplast
suspensions.
He added different concentrations of each chemical to illuminated chloroplast
suspensions containing DCPIP. He then determined the IC50 for each chemical.
The IC50 is the concentration of chemical which inhibits the decolourisation of
DCPIP by 50%.
Explain the advantage of the student using the IC50 in this investigation.
[1 mark]

Answer 28

Provides a standard / reference point
OR
Can compare different chemicals/weedkillers
OR
Can compare different concentrations of
chemicals/weed-killer

Provides a standard / reference point
OR
Can compare different chemicals/weedkillers
OR
Can compare different concentrations of
chemicals/weed-killer

Slightly more than 40% of students obtained this mark, often by stating that different
chemicals or concentrations of chemicals could be compared as weed killers. Fewer
students referred to using IC50 as a standard or benchmark. Responses which did not gain
credit often simply stated ‘for comparison’, without qualification, or simply repeated the
information provided in the stem of the question.

Question 29

0 4 A student isolated chloroplasts from spinach leaves into a solution to form a
chloroplast suspension. He used the chloroplast suspension and DCPIP solution
to investigate the light-dependent reaction of photosynthesis. DCPIP solution is
blue when oxidised and colourless when reduced.
The student set up three test tubes as follows:
• Tube 1 – 1 cm3 of solution without chloroplasts and 9 cm3 of DCPIP solution in
light.
• Tube 2 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in
darkness.
• Tube 3 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in light.
The student recorded the colour of the DCPIP in each of the tubes at the start and
after the tubes had been left at 20 °C for 30 minutes.
His results are shown in Table 1.
Table 1
Tube
Colour of DCPIP in tube
At start After 30 minutes
1 blue blue
2 blue blue
3 blue colourless
0 4 . 1 The solution that the student used to produce the chloroplast suspension had the
same water potential as the chloroplasts.

0 4 . 5 Explain how chemicals which inhibit the decolourisation of DCPIP could slow the
growth of weeds.
[2 marks]

Answer 29

1. Less/no ATP produced;
2. Less/no reduced NADP produced;
3. Less/no GP reduced/converted to TP;

Many students provided responses which showed understanding that the light-dependent
reaction would not take place, and hence neither would the Calvin cycle. This was then
often linked to less glucose being produced for growth. However, many of these answers
failed to mention the products of the light-dependent reaction. Consequently, almost 50%
of students scored zero. Better answers did refer to ATP and/or reduced NADP not being
produced for use in the light-independent reaction. Relatively few students specified that
the reduction of GP to form TP would not take place.

Question 30

Q1- TOPIC test

Using figure 1 explain the results obtained in figure 2 (2)

Answer 30

1. More o2 there is the less likely RUBP will react with cO2

2. Competitive inhibition/ competition between o2 and co2 for rubisco

Question 31

(c) Use the infomation provided and your knowledge of the LIR to explain why the yield from soya bean plants is decreased at higher concentrations of o2. Phosphoglycerate is not used in the light independent reaction (3)

Answer 31

1. Less GP produced
2. Less triose phosphate to form sugars
3. less RUBP regenerated

Question 32

Describe the role of electron transport chains in the light-dependent reactions of photosynthesis (6)

Answer 32

1. ETC accepts excited electrons
2. From chlorophyll/photosystem
3. Electrons lose energy along chain
4. ATP produced
5. from ADP and PI
6. Reduced NADP formed
7. when electrons (from transport chain) and H+ combine with NADP
8. H+ from photolysis of water

Question 33

2c- Explain why the increase in dry mass of a plant over 12 months is less than the mass of hexose produced over the same period (3)

Answer 33

• some hexose used in respiration
• co2 produced
• some parts of the plant are eaten

Question 34

4c- ATP and reduced NADP are two products of the light dependent reactions.

Describe 1 function of each of these substances in the light-independent reactions (2)

Answer 34

1. ATP provides energy for reduction of GP to TP / provides P for RUP
2. Reduced NADP provides electrons for GP to TP