Your answers will be automatically graded when you submit.
Question Navigation
This paper covers the whole of Topic 6. Like a real Cambridge paper, the seven questions range across every sub-topic — photosynthesis and leaf structure — and they are deliberately mixed rather than grouped. All three Topic 6 papers do; they differ in the angle they come at it from, not in what they cover.
Question 1 -- Counting Bubbles From Elodea
Total: 12 marks
A student placed a 6 cm length of Elodea in a boiling tube of water containing sodium hydrogencarbonate solution at 20 °C, and counted the bubbles released in one minute at five distances from a lamp. Each reading was taken three times.
distance / cm
10
20
30
40
50
reading 1
46
22
11
6
4
reading 2
50
19
13
19
4
reading 3
48
22
12
6
3
mean
48
21
12
?
3.7
(a)[3]
Complete the table by calculating the mean at 40 cm, explaining how you have treated the readings.
Model Answer -- 1(a)
reading 2 at 40 cm (19) is anomalous: it does not fit the pattern and is larger than the value at 30 cm [1]
it should be excluded from the mean, and ideally that reading repeated [1]
mean of the remaining values = (6 + 6) ÷ 2 = 6.0 bubbles per minute [1]
⚠ If you missed marks here: Including the anomaly gives a mean of 10.3, which is higher than the value at 30 cm and makes the trend nonsense. Anomalies are identified, explained and excluded — never silently deleted and never quietly averaged in.
(b)[3]
Describe the relationship between distance from the lamp and the rate of bubbling, using figures from the table.
Model Answer -- 1(b)
as the distance increases, the rate of bubbling decreases [1]
the decrease is not proportional: doubling the distance from 10 to 20 cm reduces the mean from 48 to 21, roughly a quarter, not a half [1]
the curve becomes much shallower at greater distances, falling only from 6.0 to 3.7 between 40 and 50 cm [1]
⚠ If you missed marks here: “It goes down” is worth at most one of the three marks. A describe instruction with a table in front of you always means quote the numbers, and here the shape of the fall is the interesting part.
(c)[3]
Explain why light intensity is not a good thing to represent by distance on the x-axis of a graph.
Model Answer -- 1(c)
light intensity is proportional to 1 ÷ distance², so equal steps in distance are not equal steps in intensity [1]
this makes the x-axis non-linear, so the shape of the graph does not show the true relationship between intensity and rate [1]
a light meter should be used instead, and the room darkened so the lamp is the only source [1]
⚠ If you missed marks here: Do not answer “because the lamp gets hot” here — that is a genuine problem but it is a control of variables issue, not a problem with the axis. Read exactly what the question is asking about.
(d)[3]
Suggest why the student added sodium hydrogencarbonate solution to the water, and state what would have happened without it.
Model Answer -- 1(d)
sodium hydrogencarbonate releases carbon dioxide into the water, keeping the supply of raw material high [1]
without it, carbon dioxide would become the limiting factor at the higher light intensities [1]
so the rate would plateau at short distances and the investigation would no longer be measuring the effect of light [1]
⚠ If you missed marks here: Soda lime does the opposite of this and would abolish photosynthesis altogether. The two chemicals are the most confused pair in the topic, and getting them the wrong way round inverts every prediction that follows.
Question 2 -- Sun Leaves and Shade Leaves
Total: 12 marks
Leaves were collected from the top of a beech tree, in full sun, and from a low branch in deep shade inside the canopy. Ten leaves of each type were measured.
measurement
sun leaf
shade leaf
mean area / cm²
19
44
mean thickness / µm
310
145
number of palisade layers
2
1
stomata per mm² (lower surface)
340
190
chlorophyll per unit area / arbitrary units
72
95
(a)[3]
Describe two differences between the sun leaf and the shade leaf, using figures from the table.
Model Answer -- 2(a)
the shade leaf has a much larger mean area, 44 cm² against 19 cm², more than twice as large [1]
the shade leaf is less than half as thick, 145 µm against 310 µm, and has one palisade layer instead of two [1]
any further difference quoted with figures, for example 190 against 340 stomata per mm² [1]
⚠ If you missed marks here: Marks here are for comparative statements with numbers. Two separate sentences that describe each leaf on its own often score nothing, because the command word was describe two differences.
(b)[3]
Suggest why the shade leaf has a larger area but fewer stomata per mm².
Model Answer -- 2(b)
in deep shade light is the limiting factor, so a larger surface area intercepts more of the little light available [1]
the rate of photosynthesis is therefore lower, so the demand for carbon dioxide is lower [1]
fewer stomata are needed to supply that demand, and having fewer also reduces water loss [1]
⚠ If you missed marks here: “It needs fewer stomata because it is in the shade” states the correlation without the mechanism. The chain the examiner wants is light limiting → lower rate → lower CO₂ demand → fewer pores.
(c)[3]
Explain the advantage to the sun leaf of having two palisade layers.
Model Answer -- 2(c)
two layers give more chloroplasts beneath each unit of surface area [1]
in full sun enough light penetrates to reach the second layer, so those extra chloroplasts are used rather than wasted [1]
the rate of photosynthesis per unit area is therefore higher than it could be with a single layer [1]
⚠ If you missed marks here: A second palisade layer would be a liability in shade, because the cells would cost energy to build and maintain but receive too little light to repay it. The best answers say why the adaptation only pays off in one environment.
(d)[3]
The shade leaf contains more chlorophyll per unit area than the sun leaf. Suggest an explanation.
Model Answer -- 2(d)
more chlorophyll per unit area allows a greater proportion of the scarce light to be absorbed rather than passing straight through the leaf [1]
the chloroplasts are concentrated in a single palisade layer close to the surface, so almost all of them lie in the path of the light that does arrive [1]
this is the same strategy as the larger area — maximise capture of whichever factor is limiting [1]
⚠ If you missed marks here: Answers that say the shade leaf is “greener because it is healthier” miss the logic entirely. Every difference in this table is an investment decision about the factor that is in shortest supply.
Question 3 -- Carbon Dioxide in a Sealed Glasshouse
Total: 12 marks
The carbon dioxide concentration of the air in a sealed glasshouse of tomato plants was recorded every four hours through one summer day. Sunrise was at 06:00 and sunset at 19:00.
time
00:00
04:00
08:00
12:00
16:00
20:00
CO₂ / %
0.049
0.052
0.036
0.019
0.022
0.043
(a)[3]
Describe the pattern shown by the data.
Model Answer -- 3(a)
the concentration rises to a maximum of 0.052 % at 04:00 [1]
it then falls steeply to a minimum of 0.019 % at 12:00, less than half the value eight hours earlier [1]
from 12:00 it rises again, reaching 0.043 % by 20:00, so the value at the end of the day is approaching that at the start [1]
⚠ If you missed marks here: One mark at most for “it goes down then up”. Quote the maximum, the minimum and the times at which they occur — those are the three separate marking points.
(b)[3]
Explain the value recorded at 04:00.
Model Answer -- 3(b)
at 04:00 it is dark, so no photosynthesis is occurring [1]
the plants have been respiring throughout the night, releasing carbon dioxide into the sealed air [1]
with nothing removing it, the concentration has risen well above the 0.04 % of ordinary air [1]
⚠ If you missed marks here: The sealed glasshouse is essential to the explanation: in open air the excess would simply diffuse away. Whenever a question specifies a closed system, that detail is usually load-bearing.
(c)[3]
Between 06:00 and 08:00 the concentration passes through 0.040 %. State the name given to the point at which the rate of photosynthesis equals the rate of respiration, and explain what is happening to the gases at that moment.
Model Answer -- 3(c)
the compensation point [1]
carbon dioxide is being used in photosynthesis exactly as fast as it is being produced by respiration [1]
so there is no net change in the concentration of carbon dioxide, and equally no net exchange of oxygen [1]
⚠ If you missed marks here: “Nothing is happening” is the answer to avoid. Both processes are running at full speed; they simply cancel. Distinguishing zero rate from zero net change is the whole point of the term.
(d)[3]
Explain what the value at 12:00 tells the grower, and suggest what she should do.
Model Answer -- 3(d)
by 12:00 the concentration has fallen to 0.019 %, less than half the atmospheric value, so carbon dioxide has become the limiting factor [1]
increasing light or temperature would therefore not raise the rate any further [1]
she should enrich the air with carbon dioxide from a cylinder or a burner, or ventilate the glasshouse so that outside air at 0.04 % replaces the depleted air [1]
⚠ If you missed marks here: Note how much the grower has learned without touching the plants — monitoring the air alone identified the limiting factor. Answers that recommend more lighting ignore the evidence in front of them.
Question 4 -- Temperature and the Rate of Photosynthesis
Total: 12 marks
The rate of photosynthesis of a water plant was measured as the volume of oxygen collected in ten minutes, at a constant saturating light intensity and with sodium hydrogencarbonate added to the water.
temperature / °C
5
15
25
35
45
55
O₂ collected / cm³
0.4
1.1
2.6
3.1
1.2
0.0
(a)[3]
Use the data to estimate the optimum temperature for this plant and explain how you obtained your estimate.
Model Answer -- 4(a)
the optimum is between 25 and 45 °C, close to 35 °C where the highest volume of 3.1 cm³ was collected [1]
the true optimum could lie anywhere between the tested values, so readings at 5 °C intervals between 25 and 45 °C would be needed to locate it precisely [1]
each reading should also be repeated and a mean taken, so that the peak is not an artefact of a single anomalous measurement [1]
⚠ If you missed marks here: Stating “35 °C” with no qualification gains one mark at most. The second mark is for recognising that a tested value is not necessarily the peak — the same interpolation thinking you use for an isotonic point.
(b)[3]
Explain the shape of the graph between 5 and 35 °C.
Model Answer -- 4(b)
as temperature rises the enzyme and substrate molecules gain kinetic energy and move faster [1]
so they collide more frequently and with more energy, forming more enzyme–substrate complexes per second [1]
therefore the rate of the reactions of photosynthesis, and so the volume of oxygen released, increases [1]
⚠ If you missed marks here: Answers about carbon dioxide dissolving better or the plant “working harder” do not score. Temperature acts on the enzymes, and the mark scheme is written in the language of kinetic energy and collisions.
(c)[3]
Explain the shape of the graph above 35 °C.
Model Answer -- 4(c)
above the optimum the enzymes are denatured [1]
the high temperature breaks the bonds holding the protein in its precise shape, so the active site changes shape [1]
the substrate no longer fits, so fewer enzyme–substrate complexes form and by 55 °C the rate has fallen to zero [1]
⚠ If you missed marks here: “The enzymes are killed” is refused because enzymes are molecules, not organisms. This is also why the temperature curve falls where light and carbon dioxide curves merely plateau — a plateau is a change of limiting factor, a fall is damage.
(d)[3]
Explain why the light intensity was kept saturating and sodium hydrogencarbonate was added throughout.
Model Answer -- 4(d)
so that neither light intensity nor carbon dioxide concentration became the limiting factor at any temperature [1]
this means that any change in the rate can be attributed to the temperature alone, making the investigation valid [1]
without them the curve would plateau for the wrong reason and the true effect of temperature would be hidden [1]
⚠ If you missed marks here: This is a controlled-variables mark dressed up in the language of limiting factors, and it is a good example of why the two ideas are really the same idea seen from different sides.
Question 5 -- Starch in a Leaf Over 24 Hours
Total: 10 marks
Discs of equal area were cut from the leaves of a bean plant every four hours for one day and the mass of starch in each was measured. The plant was in natural daylight, with sunrise at 06:00 and sunset at 18:30.
time
02:00
06:00
10:00
14:00
18:00
22:00
starch / mg per disc
0.9
0.4
1.8
3.4
4.1
2.2
(a)[3]
Explain why the starch content rises between 06:00 and 18:00.
Model Answer -- 5(a)
in daylight the leaf photosynthesises, producing glucose [1]
glucose is produced faster than it can be used in respiration or transported away, and the surplus is converted to starch and stored [1]
starch is insoluble, so it can accumulate in the cell without affecting water potential or leaking out [1]
⚠ If you missed marks here: “The plant makes starch in photosynthesis” is not accurate: photosynthesis makes glucose, and starch is made from it afterwards. That distinction is worth a mark on almost every paper in this topic.
(b)[3]
Explain why the starch content falls between 18:00 and 06:00.
Model Answer -- 5(b)
after sunset there is no light, so no photosynthesis and no new glucose is made [1]
the stored starch is converted back to glucose for respiration, and to sucrose for transport in the phloem to the roots and other organs [1]
so the store is drawn down until dawn, when it reaches its minimum of 0.4 mg per disc at 06:00 [1]
⚠ If you missed marks here: Answers that say the starch “diffuses out of the leaf” are wrong: starch is far too large and is insoluble. It must be converted to a soluble form first, which is exactly why sucrose exists.
(c)[2]
Discs of equal area were used rather than whole leaves. Explain why.
Model Answer -- 5(c)
so that the results can be compared fairly, as equal areas contain comparable numbers of cells [1]
leaves differ in size, age and position, so whole leaves would not be a fair comparison [1]
⚠ If you missed marks here: This is a standardisation mark, closely related to the idea of a controlled variable. The same reasoning is behind measuring stomata per mm² rather than per leaf.
(d)[2]
Suggest why the value at 06:00 is not zero.
Model Answer -- 5(d)
conversion and export of starch take time, so the store is not completely emptied in one night [1]
the plant retains a reserve so that respiration can continue if the following day is dull, and a leaf destarched fully requires 24 to 48 hours of darkness [1]
⚠ If you missed marks here: This part quietly explains why destarching requires two days rather than one night, which is the detail most candidates get wrong when describing photosynthesis experiments.
Question 6 -- Where Are the Stomata?
Total: 12 marks
Four leaves of equal area were taken from one plant and treated with petroleum jelly, which is waterproof and blocks stomata. Each leaf was hung in the same room and weighed every hour for three hours; the loss of mass is almost entirely loss of water.
leaf
treatment
mass lost in 3 h / g
W
none
0.84
X
upper surface coated
0.79
Y
lower surface coated
0.16
Z
both surfaces coated
0.03
(a)[3]
Use the data to deduce the distribution of stomata on this leaf, showing how the figures support your conclusion.
Model Answer -- 6(a)
coating the lower surface reduced water loss from 0.84 to 0.16 g, a fall of about 80 % [1]
coating the upper surface reduced it only from 0.84 to 0.79 g, a fall of about 6 % [1]
therefore the great majority of the stomata are on the lower surface [1]
⚠ If you missed marks here: The conclusion alone is worth one mark of three. Deduction questions require you to show the reasoning from the figures, and percentages make the contrast much clearer than raw masses.
(b)[3]
Explain why leaf Z still lost 0.03 g of water.
Model Answer -- 6(b)
a small amount of water evaporates through the waxy cuticle itself, which greatly reduces but does not abolish water loss [1]
the loss is only 0.03 g against 0.84 g for the untreated leaf, about 4 %, showing how effective a barrier the cuticle is [1]
some loss may also occur from the cut petiole or from surfaces the jelly did not completely cover [1]
⚠ If you missed marks here: Do not answer “some stomata inside the leaf were still open”. There are no stomata inside a leaf; the pores are in the epidermis, which is exactly what was coated.
(c)[3]
Explain the advantage to the plant of having most of its stomata on the lower surface, referring to both of the leaf’s competing requirements.
Model Answer -- 6(c)
the plant must allow carbon dioxide to diffuse in for photosynthesis, which requires open pores [1]
but open pores also allow water vapour to be lost from the moist surfaces of the air spaces [1]
the lower surface is shaded and cooler and sheltered from air movement, so the rate of evaporation there is lower while gas exchange still occurs [1]
⚠ If you missed marks here: The word the examiner is looking for is a compromise or trade-off. Answers that mention only water loss, or only carbon dioxide, describe half of the problem the leaf has to solve.
(d)[3]
Suggest one change to this method that would make the results more reliable, and one that would make them more valid.
Model Answer -- 6(d)
reliability: repeat the whole investigation with several leaves given each treatment and calculate a mean, so that random variation between individual leaves is reduced [1]
validity: use leaves of the same age and area from the same plant, and keep temperature, humidity and air movement identical for all four leaves [1]
validity: seal the cut end of each petiole, so that all the mass lost is lost through the leaf surfaces being tested [1]
⚠ If you missed marks here: Reliability and validity are not interchangeable. Repeats deal with random error; controls deal with whether you measured what you claimed to measure. Read which word the question uses.
Question 7 -- Magnesium and Chlorophyll
Total: 10 marks
Young maize plants were grown in culture solutions containing different concentrations of magnesium ions, with all other ions supplied in full. After four weeks the chlorophyll content of the leaves and the dry mass of each plant were measured.
Mg²⁺ / mg dm⁻³
0
5
10
20
40
80
chlorophyll / arbitrary units
4
27
49
71
74
75
dry mass / g
0.21
0.58
0.96
1.34
1.39
1.40
(a)[2]
Describe the relationship between magnesium concentration and chlorophyll content.
Model Answer -- 7(a)
as the magnesium concentration increases, the chlorophyll content increases, from 4 units at 0 to 71 units at 20 mg dm⁻³ [1]
above 20 mg dm⁻³ the chlorophyll content levels off at around 75 units, so further magnesium has little effect [1]
⚠ If you missed marks here: Two marks, two features: the rise and the plateau. Describing only the rise is the usual half-answer, and the plateau is the more interesting half.
(b)[3]
Explain why plants grown without magnesium had so little chlorophyll and such a low dry mass.
Model Answer -- 7(b)
magnesium is required to make chlorophyll, so without it almost no chlorophyll can be synthesised [1]
without chlorophyll, little light energy can be transferred into chemical energy, so the rate of photosynthesis is very low [1]
little glucose is made, so little material is built into the plant and the dry mass stays low at 0.21 g [1]
⚠ If you missed marks here: The chain must be complete: magnesium → chlorophyll → energy transfer → glucose → dry mass. Jumping straight from “no magnesium” to “the plant is small” skips three marking points.
(c)[3]
Explain why adding magnesium above 20 mg dm⁻³ produced almost no further increase in dry mass.
Model Answer -- 7(c)
above 20 mg dm⁻³ magnesium is no longer in short supply, so it is no longer the factor limiting chlorophyll production or growth [1]
some other factor — light intensity, carbon dioxide concentration or temperature — has become the limiting factor [1]
so supplying more magnesium cannot raise the rate any further [1]
⚠ If you missed marks here: This is limiting-factor reasoning applied to a mineral ion rather than to light or carbon dioxide, and the logic is identical: increasing a factor that is not limiting changes nothing.
(d)[2]
Magnesium ions are taken into root hair cells even when the soil solution is far more dilute than the cell contents. Name the process involved and explain how it is possible.
Model Answer -- 7(d)
the process is active transport, because the ions move against the concentration gradient [1]
protein carriers in the cell surface membrane move the ions across, using energy released by aerobic respiration in the mitochondria of the root hair cell [1]
⚠ If you missed marks here: Never decide the process from the identity of the particle. It is the direction of the gradient that determines whether something moves by diffusion or by active transport, and here the gradient runs the wrong way.
Self-Assessment
Tick marks earned, then click Calculate Grade.
0
80
0%
A* : 56+
A : 48-55
B : 40-47
C : 32-39
D : 24-31
E : 16-23
U : <16
When finished, click Submit to see model answers.
Exam Submitted -- Marking Mode Active
Click "Show Model Answer" on each question to check your work.