Author: Philip Colls
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Mass transport – Plants booklet revision questions      NAME:
Q1.
Environmental factors can affect the density of stomata in the lower epidermis of leaves of plants of the same species.
Scientists investigated how growing plants at different temperatures affected the density of stomata in the lower epidermis of leaves. They grew plants of the same species from seeds.
Their method is outlined below.
•        They took 8 trays containing soil and planted 50 seeds in each tray.
•        They put each tray in a controlled environment at a different temperature.
•        When the plants had grown from the seeds, they selected 20 fully grown leaves from the plants in each tray.
•        They determined the mean number of stomata per mm^2 in the lower epidermis for each group of leaves.
Their results are shown in the graph.
 
(a)     Give three environmental variables, other than temperature, that the scientists would have controlled when growing the plants.
1. _________________________________________________________________
2. _________________________________________________________________
3. _________________________________________________________________
(3)
(b)     The scientists used a range of temperatures from 6 to 20 °C.
Using their data, explain why they did not use temperatures above 20 °C.
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(1)
(c)     The scientists only selected fully grown leaves from the plants.
Suggest why.
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(1)
(d)     The plants grown at higher temperatures had a lower number of stomata per mm^2.
This would be an advantage to the plant because the transpiration rate increases as the temperature increases.
Explain why the transpiration rate increases when the temperature increases.
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(2)
(Total 7 marks)
Q2. The drawing shows four common plants found in the Mojave Desert.
W                   X                   Y                   Z
 
(a)     Explain how three features of the plants shown in the drawing are adaptations to desert conditions.
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2. _________________________________________________________________
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3. _________________________________________________________________
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(b)     Resurrection plants can lose up to 95% of their water content without dying. They can survive for many years in this desiccated state and will revive within hours of rainfall. Suggest which of the plants W to Z is most likely to be a resurrection plant. Give a reason for your choice.
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Q3. (a)     The table shows the transpiration rate of a group of plants exposed to different humidities at a temperature of 25°C.

Humidity / %

Transpiration rate /
arbitrary units

20 | 26.0
40 | 21.0
50 | 16.5
60 | 11.0
70 | 9.5
Describe and explain the relationship between humidity and transpiration rate.
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(b)     The diagrams show a section through a typical leaf and a section through a leaf from a xerophytic plant. The xerophytic leaf has a lower transpiration rate than the typical leaf.

Describe two features shown in the diagram of the xerophytic leaf which reduce transpiration rate. Explain how each of these features contributes to a lower transpiration rate.
Feature 1___________________________________________________________
Explanation _________________________________________________________
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Feature 2___________________________________________________________
Explanation _________________________________________________________
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(4)
(Total 7 marks)

Q4. (a)     Describe how water is moved through a plant according to the cohesion-tension hypothesis.
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(4)
(b)     The mass of water lost from a plant was investigated. The same plant was used in every treatment and the plant was subjected to identical environmental conditions. In some treatments, the leaves were coated with a type of grease. This grease provides a waterproof barrier. The results of the investigation are given in the table.
 
Treatment | Mass lost in 5 days / g
No grease applied | 10.0
Grease applied only to the upper surface of every leaf | 8.7
Grease applied to both surfaces of every leaf | 0.1
(i)      What is the advantage of using the same plant in every treatment?
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(1)
(ii)     Why was it important to keep the environmental conditions constant?
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(1)
(iii)     What is the evidence that the grease provides a waterproof barrier?
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(1)
(c)     (i)      Calculate the mass of water lost in 5 days through the upper surface of the leaves.

 
Answer __________________________________(1)
(ii)     Use your knowledge of leaf structure to explain why less water is lost through the upper surface of leaves than is lost through the lower surface.
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(Total 10 marks)
Q5.
(a)     Describe the mass flow hypothesis for the mechanism of translocation in plants.
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________________________________________________________________ (4)
Scientists measured translocation in the phloem of trees. They used carbon dioxide labelled with radioactive ^14C.
They put a large, clear plastic bag over the leaves and branches of each tree and added ^14CO_2. The main trunk of the tree was not in the plastic bag.
At regular intervals after adding the ^14CO_2 to the bag, the scientists measured the amount of ^14CO_2 released from the top and bottom of the main trunk of the tree. On the surface of the trunk of these trees, there are pores for gas exchange.
The following figure shows the scientists’ results.
 
                               Time after ^14C labelled CO_2 given / hours
(b)     Name the process that produced the ^14CO_2 released from the trunk.
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(1)
(c)     How long did it take the ^14C label to get from the top of the trunk to the bottom of the trunk? Explain how you reached your answer.
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(2)
(d)     What other information is required in order to calculate the mean rate of movement of the ^14C down the trunk?
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(1)
(Total 8 marks)
Q6. Under the correct conditions, new roots grow from the cut end of a plant stem. A scientist investigated the effect of substance X on the growth of new roots.
She used a ringing experiment to investigate the movement of substance X in stems taken from lemon plants. She cut out a length of stem from each plant. She then put a small block of agar on the top of each length of stem. Some agar blocks contained substance X.
The diagram below shows how she treated each length of stem.

She grew the lengths of stem in the same environmental conditions for 6 weeks, and then found the number of roots per length of stem. Roots grew at the other end of the stem from where the agar blocks were placed.
The table below shows the scientist’s results.
 
Treatment | Mean number of roots per length of stem
D | 5
E | 11
F | 4
G | 3
(a)     Treatment D is a control. Explain how the measurement obtained from this control is used by the scientist.
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(b)     Using the diagram and the table above, what can you conclude from treatments D and E about root growth?
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(3)
(c)     The mass flow hypothesis is used to explain the movement of substances through phloem.
Evaluate whether the information from this investigation supports this hypothesis.
Do not consider statistical analysis in the answer.
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 (4)
(Total 9 marks)
Q7. A student used a potometer to measure the movement of water through the shoot of a plant. The potometer is shown in Figure 1. As water is lost from the shoot, it is replaced by water from the capillary tube.
 
(a)     In one experiment, the air bubble moved 7.5 mm in 15 minutes. The diameter of the capillary tube was 1.0 mm. Calculate the rate of water uptake by the shoot in this experiment.
Give your answer in mm^3 per hour. Show your working. (The area of a circle is found using the formula, area = πr ^2)
____________________ mm^3 hour^−1
(2)
(b)     The student wanted to determine the rate of water loss per mm^2 of surface area of the leaves of the shoot in Figure 1.
Outline a method she could have used to find this rate. You should assume that all water loss from the shoot is from the leaves.
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 (3)
(c)     The rate of water movement through a shoot in a potometer may not be the same as the rate of water movement through the shoot of a whole plant.
Suggest one reason why.
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(d)     Aquaporins are channel proteins that allow the diffusion of water across membranes. One type of aquaporin, called PIP1, can also transport carbon dioxide molecules across membranes.
Figure 2 shows the structure of a water molecule and of a carbon dioxide molecule. They are drawn to the same scale.
 
Suggest two reasons why water molecules and carbon dioxide molecules can both pass through PIP1.
1. _________________________________________________________________
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2. _________________________________________________________________
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(2)
(e)     The scientists first produced transgenic poplar trees. These trees all had a length of foreign DNA inserted into them. This DNA led to the production of single-stranded RNA that specifically inhibited expression of the gene for PIP1. The scientists then measured the difference in the amount of PIP1 in leaves of transgenic poplars and in leaves of wild type poplars without the foreign DNA. The amount of PIP1 in the transgenic poplars was approximately 15% of that in the wild type poplars.
Using this information, what can you conclude about the effect of the foreign DNA in the transgenic poplar trees?
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(f)      The transgenic poplars still produced some PIP1. Suggest why.
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(g)     The scientists investigated the importance of PIP1 in the movement of water and carbon dioxide through the tissues of leaves of poplar trees. They measured the mean rates of movement of carbon dioxide and water through the tissues of leaves of transgenic poplars and through the tissues of leaves of wild type poplars.
Their results are shown in the graph below.
 
Using only the graph above, evaluate the importance of PIP1 in the movement of carbon dioxide and water through leaves of poplar trees.
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(3)
(Total 15 marks)

Mark schemes
Q1.
(a)     Any three from:
1.      Light;
2.      Carbon dioxide;
3.      Type of soil;
4.      Minerals / nutrients;
Accept named example
5.      Water (in soil);
6.      Humidity (of air);
7.      pH (of soil)
8.      Planting density;
Idea of equally spaced
3 max
(b)     Already levelled out (before 20 °C);
1
(c)     Young leaves (may) have different number of stomata (per mm^2) / number of stomata (per mm^2) changes during development (of leaf);
Accept reference to density of stomata
1
(d)     Any two from:
Points 1 and 2 need context of ‘more’
1.      Molecules have more kinetic energy;
Accept KE
2.      Faster diffusion of water / more evaporation of water (as temperature increases in leaf);
For this point, diffusion must relate to movement of water
3.      For this point, diffusion must relate to movement of water
2 max
[7]
Q2.
(a)     shallow roots enable rapid uptake of rainfall (in X and / or Z);
widespread / shallow roots allow collection of larger volume
water / over a larger area / rapid uptake of water (in Z);
swollen stem for water storage (in X);
deep roots for accessing deep groundwater (in Y);
small / no leaves so little transpiration;
3
(b)     Z;
wide spread of roots for rapid water absorption;
(accept X; if linked to leaves channelling water to roots)
(ignore references to water storage abilities)
(accept other responses if justified)
2
[5]
Q3.
(a)     increased humidity leads to decreased transpiration;
high humidity means more water in the air / increased saturation / 
increased water potential;
reduced diffusion gradient / water potential gradient;
slower rate of water loss / less evaporation;
3 max
(b)     thick cuticle; 
impermeable to water / waterproof;
sunken stomata; 
reduces water diffusion gradient;
shape of leaf / rounded / small surface area;
small surface area : volume ratio;
(explanation must be linked to feature)
4 max
[7]
Q4.
(a)     1. water evaporates / transpires from leaves;
2. reduces water potential in cell / water potential / osmotic gradient across
    cells (ignore reference to air space);
3. water is drawn out of xylem;
4. creates tension (accept negative pressure, not reduced pressure);
5. cohesive forces between water molecules;
6. water pulled up as a column;
4 max
(b)     (i)      same surface area of leaf / number of leaves / age / thickness of
cuticle;
1
(ii)     (environmental conditions) affect rate of transpiration / evaporation;
1
(iii)     presence of grease reduces water loss;
1
(c)     (i)      1.2 / 1.3g;
1
(ii)     more stomata on the lower surface;
(thicker) waxy cuticle on the upper surface;
2
[10]
Q5.
(a)     1.      In source / leaf sugars actively transported into phloem;
2.      By companion cells;
3.      Lowers water potential of sieve cell / tube and water enters by osmosis;
4.      Increase in pressure causes mass movement (towards sink / root);
5.      Sugars used / converted in root for respiration for storage.
Accept starch
4 max
(b)     Respiration.
1
(c)     1.      (About) 30 hours;
2.      Time between peak ^14C at top of trunk and bottom.
2
(d)     Length of trunk (between top and bottom).
1
[8]
Q6.
(a)     1.      Used to compare effect of other treatments / as a baseline;
Accept for 2 marks, substance (X) and not agar / block / water that caused the difference in the number of roots.
Do not accept unqualified reference to “compare results”.
2.      Shows / Measures effect of substance (X);
OR
Accounts for effect of substances produced naturally;
Accept measures effect of independent variable
2
(b)     1.      (D shows) substance (X) is not required for (some) root growth / production of roots;
OR
Substances (already) present in stem cause (some) root growth;
2.      Substance X moves through plant;
Accept X moves through stem / phloem
3.      (E shows) substance (X) causes / increases / doubles number of roots / root growth;
3
(c)     In support of mass flow hypothesis
1.      (F shows) phloem is involved;
2.      (G shows) respiration / active transport is involved (in flow / movement);
3.      Because 4 °C / cooling reduces / slows / stops flow / movement;
4.      The agar block is the source;
5.      Roots are the sink;
Against the mass flow hypothesis
6.      No bulge above ringing (in F);
7.      No (role for) osmosis / hydrostatic pressure / water movement;
Accept no turgor pressure
8.      Movement could be due to gravity;
9.      Roots still grow without (intact/functioning) phloem;
10.    No leaves / sugars / photosynthesis to act as a source;
Each point must be clearly made in the context of support or against.
Ignore sugar / sucrose
3 max for “support” and 3 max for “against”
4 max
[9]
Q7.
(a)     Correct answer 23.55 – 24 two marks;
For one mark
5.9
OR
94.2;
2
(b)     1.      Method for measuring area;
e.g. draw round (each) leaf on graph paper and count squares;
2.      Of both sides of (each) leaf;
3.      Divide rate (of water loss / uptake from potometer) by (total) surface area (of leaves);
3
(c)     Plant has roots
OR
xylem cells very narrow;
Ignore references to air bubbles / mass flow / photosynthesis
Accept xylem damaged when cut
1
(d)     1.      Both small / similar size (so fit channel);
2.      Have a similar shape (so bind to / fit channel);
1.      Accept same height and width
Ignore refs to polar / non-polar
2.      Accept Aquaporin complementary to oxygen(s)
2
(e)     1.      Single-stranded RNA (has base sequence) complementary to PIP1 mRNA;
2.      Binds to mRNA (of PIP1) / leads to destruction of mRNA;
3.      Prevents / reduces translation (of PIP1);
4.      Reduces photosynthesis/named process that uses water;
3.      Less made is insufficient
3 max
(f)      Not all of mRNA bound to single-stranded RNA / there is more mRNA than interfering RNA
OR
Not all mRNA destroyed / disabled;
Accept mutations in transgene,
Accept not all cells with transgenes
1
(g)     1.      Loss of PIP reduces water and carbon dioxide movement;
2.      Differences significant because SDs don’t overlap
OR
Need stats test to see whether significant differences (or not);
3.      Greater (proportional) effect on carbon dioxide transport;
4.      Not all movement through PIP;
1.      Accept converse for wild type
2.      Reject references to results significant or not significant
2.      Accept error bars for SDs
3 max
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