B io Factsheet
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Number 225
Synoptic Biology: Water Potential
This Factsheet reviews the very wide range of exam questions that test your understanding of water potential.
Osmosis: the diffusion of water through a partially permeable membrane from a region with more free water to a region with less free water
/ down a water concentration gradient.
Key Point
Water moves from a region of high water potential to low water
potential. This fact gives us the simplest type of exam question.
Typical Exam Question 2
Fig.2 shows the water potential of four neighbouring spongy
mesophyll cells from the leaf of a daffodil plant.
Fig 2
Typical Exam Question 1
Fig. 1 shows the water potential of four plant root cells.
Ψ = -850
Ψ = -900
Ψ = -1050
Cell A
Ψ = -1700 kPa
Cell B
Ψ = -1300 kPa
Ψ = -1300
Draw arrows on Fig.2 to show the net flow of water between the
cells (3).
Cell D
Ψ = -1200 kPa
Water will always move towards the most negative number. So hopefully,
you drew arrows as follows:
arrow drawn from -850 to -900 , from -850 to -1050 and from -1050 to
-1300
Cell C
Ψ = -1975 kPa
Each cell was placed into a sucrose solution with a water potential
of -540 kPa. All of the cells took in water from this solution.
Markscheme
(a) Which cell would initially take in water at the fastest rate?
Explain why (3).
Exam Hint:- Did you draw a continuous arrow from -850 to 1050 to –1300 ? If you did, you understand water potential but
your exam technique needs sharpening up – 3 marks means
you have to do 3 separate things. In this case you would have
been awarded 2 out of the three.
(b) Explain why the water potential of a cell is usually a negative
value (2).
Note: In (a) 3 marks means you must make three separate points!
(a) Cell C;
It has the {lowest / most negative water potential;
Water moves from a higher to a lower water potential;
There is the steepest water potential gradient;
(b) Pure water has a water potential of zero;
Cells always contain solutes;
Solutes make the water potential lower/negative;
Extract from Chief Examiner’s Markscheme
A worrying number of candidates drew arrows both from high
to low and low to high water potential, showing a serious
misunderstanding of the concept of water potential gradients
Markscheme
Recognising plasmolysis
The third type of question involves you being able to recognise
cells that are losing water by osmosis when they are immersed in a
solution of lower water potential.
The next exam question takes this one small step further by
considering how water potential affects water movement inside the
plant.
1
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225. Synoptic Biology: Water Potential
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Fig 3 shows the appearance of a plant cell when immersed in three
different concentrations of sugar.
Table 1 shows the results of such an investigation.
Molarity of sucrose
solution/mol dm-3
Fig 3
0
0.1
0.2
0.3
0.4
0.5
cell wall cytoplasm vacuole
A
B
C
The cell has absorbed water
from the solution. The water
potential of the cell must be
lower than that of the
surrounding solution.
Initial
mass/g
Final
mass/g
1.20
1.17
1.25
1.09
1.18
1.17
1.56
1.32
1.30
1.01
1.03
0.93
% change
in mass
30.0
12.8
4.0
- 7.3
-12.7
-20.5
A graph is then drawn showing percentage change in mass against
the molarity of the bathing solution. A line of best fit is drawn and
the water potential is determined as the intersection of the line of
best fit with the x-axis.
Water has been osmotically
drawn out of the cell. The
water potential of the
solution must be less than
that of the cell.
Typical Exam Question 3
The graph shows the percentage change in the length of potato
cylinders which had been placed in sucrose solutions of different
concentrations for 12 hours.
Almost all of the water has been
osmotically drawn out of the cell
vacuole. This cell must be
immersed in the solution with
the lowest water potential. The
cell is said to be plasmolysed.
+10
+5
% change
in length
Extract from Chief Examiner’s Markscheme
0
“Some candidates continue to confuse plasmolysis and
turgidity.”
0.2
0.4
0.6
concentration
of sucrose
/mol dm-3
-5
Investigating the water potential of potato cells
This is done by immersing pieces of plant tissue e.g. potato in
different concentrations of a solution and finding the concentration
(which is equivalent to the water potential) when there is no net
gain or loss of water; the tissue remains in equilibrium with the
surrounding bathing solution.
-10
(i) In terms of water potential, explain the change in length which
occurred when the potato cylinders were placed in a sucrose
solution of concentration 0.1 mol dm-3(2).
(ii) Suggest why there was no change in mass of the potato placed
in a sucrose solution of 0.26mol dm-3 .
(iii)Explain the results obtained when the potato cylinders were
placed in sucrose solutions between 0.4 - 0.6mol dm-3(2)
(iv)The investigation was repeated using another potato and the
results were quite different. It was concluded that there was a
difference in water content of the two potatoes. Suggest three
reasons for this difference in water content (3).
(v) Suggest why measuring the change in weight of the potato
cylinders may be more accurate than measuring change in length
(2)
The effect is measured by recording changes in length or changes
in mass. By plotting the results on a graph, it is possible to estimate
the water potential of the tissue.
The usual technique that you are asked to perform or answer
questions on is as follows:
1. Use a core borer to obtain equal-sized cylinders of the potato
2. Surface-dry the cylinders using filter paper to remove any water
that is not within the plant tissue or that has resulted from cutting
3. Weigh the cylinders
(i) Water potential inside cylinders less than sucrose solution;
Water moves by osmosis into the cylinders;
(ii) No difference in water potentials;
So no net movement of water molecules;
(iii) Cell walls are rigid and prevent any further loss of water;
Potato cells remain plasmolysed;
.(iv)type / variety / genotypes / country of origin might be different;
age of potatoes may differ;
storage time;
growth/storage conditions;
part of potato used;
time to sprouting;
(v) potato cylinders may not be same width;
potato cylinders may not be straight;
change in length may be very small;
Use of rulers is subjective/, only measures changes in one plane;
5. Add the cylinders to a range of sucrose solutions of known
molarity
6. Leave for 24 hours, after which surface-dry the cylinders again
(using the same type of filter paper and drying technique/
pressure etc. as the first time) and re-weigh
7. Express the change in mass as a percentage of the initial mass to
take account of differences in initial mass
Exam Hint:- Make sure that you can explain the scientific basis
of each step.
/0 2 - 1.05/06G
Markscheme
2
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Surface area: Volume ratios
Xerophytes
The effect of the surface area: volume ratio on osmosis can be
investigated using cubes of potato.
Transpiration is inevitable. Plants need their stomata to be open for
gaseous exchange. But this simultaneously allows water to escape.
The mesophyll cells are moist to accelerate absorption of carbon
dioxide. The outside air is often drier so water diffuses out along
the water potential gradient. Xerophytes have adaptations e.g. hairs
or sunken stomata that trap moisture near the surface of the leaf
and reduce the water potential gradient, reducing, (but not stopping)
transpiration.
Typical Exam Question 4
Cubes were cut from fresh potato. The dimensions of the cubes
are shown in the diagram
1cm x 1cm x 1cm Surface area = 6cm2 Volume = 1cm3
2cm x 2cm x 2cm Surface area = 24cm2 Volume = 8cm3
Extract from Chief Examiner’s Markscheme
Candidates need to know that transpiration should be
considered in terms of loss of water vapour down water potential
gradients
2cm
1cm
1cm
1cm
2cm
Water movement across the root
2cm
The cubes were blotted dry with filter paper and weighed.
Fig.4 shows the three routes by which water moves across the cells
in the route of a dicotyledonous plant.
One cube measuring 2cm x 2cm x 2cm was placed in a beaker and
covered with distilled water
Fig 4
root hair
8 cubes, each measuring 1cm x 1cm x 1cm were put into a second
beaker and covered with distilled water
xylem
At intervals, for a period of 40 hours, the cubes were removed
from the beakers, dried, reweighed and returned to fresh beakers
of distilled water. The percentage increase in mass of all 9 cubes
was calculated and is plotted on the graph below.
Path 1 vacuolar pathway
Path 2 via cytoplasm (symplast) pathway
Path 3 via cell walls (apoplast) pathway
Common pathways
20
% increase in mass
18
16
The root hairs need to absorb water from the soil. In order to do this
they maintain low water potentials by concentrating solutes. In all
three pathways, the water is moving, via osmosis down a water
potential gradient.
14
12
10
8
Typical Exam Question 5
6
Scientists measured the water potential of soil in a field in which
barley was being successfully grown. The water potential was –
42kPa.
4
2
Which of the following water potential values is most likely to be
that of the cell sap in the root hairs of the barley plants?
0
0
10
20
30
time/hours
40
50
-38kPa
8 cubes of side 1cm x 1cm x 1cm
-42kPa
-58kPa
0kPa
Hopefully you realised that, to absorb water from the soil, the
water potential of the cell sap had to be lower – more negative –
than that of the soil solution. So, the only realistic value was –
58kPa.
1 cube of side 2cm x 2cm x 2cm
(a) Suggest why 8 cubes of dimension 1cm x 1cm x 1cm were
used in the investigation (1).
(b) Explain the results shown (2)
Water movement in xylem
Transpiration from the mesophyll cells in the leaves reduces the
water potential of the cells. Water is thus drawn out of an adjacent
cell by osmosis. As this cell loses water, its own water potential
decreases and water is drawn out of the next adjacent cell. This sets
up a water potential gradient which places water in xylem vessels
under a tension. Vessels are completely water filled so the tension,
combined with the cohesive (hydrogen bonding) and adhesive
properties of water move water up as a complete column in the
thousands of xylem vessels. To summarise, water moves up the
xylem down the water potential gradient.
Hint:-In (b) think about how surface area is likely to affect the
movement of water into the cubes.
(a) So that the total mass/volume is the same as the larger cube;
(b) Percentage increase in mass of the eight 1cm cubes was faster than
that of the 2cm cube because greater SA:V ratio/ 6:1 versus 3:1;
So water could enter by osmosis faster;
The outer cells of the larger cube may have become fully turgid;
so restricting water uptake by the inner cells;
Markscheme
3
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Seawater flooding
Potometers
We have seen that plant root hairs need to absorb water from the
soil and do this by maintaining low water potentials in their root
hairs. But what happens if the field in which they are growing
becomes flooded with seawater? The water potential of the soil
water will drop as a result of the salts in the seawater. Water will
then be drawn out of the root hairs by osmosis. This could kill the
plant very quickly.
Potometers (Fig, 6) measure the uptake of water by a shoot, they do
not measure transpiration. However, it is usually assumed that what
the plant takes up, it has lost by transpiration.
Fig. 6 Potometer
Typical Exam Question 6
leafy shoot
Explain why, after flooding with seawater, plants may wilt rapidly
(2)
Capillary tube
filled with water
reservoir
After flooding, the water potential of the soil dropped;
drawing water out of the root hairs by osmosis,
eventually causing cells throughout the plant to become flaccid;
air bubble
Markscheme
water
Mass flow in the phloem
Scale
Leaves produce sugars by photosynthesis. How does the plant get
these sugars to where they are needed e.g. a root that is actively
growing? One hypothesis suggests that plants get sugars from
sources (e.g. leaves or storage organs) to sinks (e.g. roots or other
growing regions) entirely passively (without using energy) by mass
flow. Water potential is central to this hypothesis.
Potometers can be used to investigate the variables that affect the
rate of transpiration. A typical question might provide data, either in a
table or graph that shows the effect of a fan on the rate of water
uptake. Your explanations need to focus on water potential gradients!
The rate of uptake of water by the cut shoot (which we are all assuming
is equal to the rate of transpiration) will increase when the fan is
turned on. The wind created by the fan blows away the water vapour
(don’t refer to water or water droplets). This steepens the water potential
gradient and thus more water evaporates and more water vapour
diffuses out of the stomata. Just to say that the wind causes more
evaporation without explaining why will not get any marks.
Fig. 5 Physical model to illustrate the principle of mass flow
tube
water
Typical Exam Question
1. The table shows typical values of the water potential at various
points between the soil and the top of a tree.
Source
Sink
Location
Soil
Sugar
solution
rigid, partially
permeable membranes
water
Water potential values (MPa)
-0.04
Root cells
-0.2
Stem
-0.5
Leaf cells
-1.5
Atmosphere
-98
(a) Use the information in the table to outline how water may be
drawn from the soil to the top of a tree. (3)
How it works:
1. The sugars produced via photosynthesis in leaves or hydrolysis
of starch in roots lowers the water potential of cells there
2. So water enters the source by osmosis
3. Down the water potential gradient
4. This results in an increase in hydrostatic pressure because the
source cannot expand
5. This forces the solution along the tube to the sink
6. As a mass flow
(b) State 3 environmental factors which affect the rate of
transpiration. (3)
Semicolons indicate marking points
1. (a) Water moves from a region of high water potential to a region of
low water potential;
water potential gradient exists;
favouring movement of water from soil to root / root to stem /
stem to leaf cell / leaf to air; max 3
(b) atmospheric humidity;
temperature;
internal CO2 concentration;
water solubility;
light intensity; max 3
Extract from Chief Examiner’s Markscheme
Many candidates muddled sources and sinks, perhaps because
roots can be both! Too many candidates seemed to believe that
mass flow is the diffusion of sugar from the source to the sink
down a concentration gradient.
Markscheme
4
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225. Synoptic Biology: Water Potential
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Stomatal opening and closing
Practice Questions
Abscisic acid (ABA) is a plant growth regulator synthesised in the
leaves. It has been suggested that plants synthesise ABA when
they are facing water stress, in order to cause the stomata to close.
1. The diagram represents three plant cells and shows the water
potential of two of these cells. The arrows show the direction of
water movement between these three cells.
A
In an investigation into the effects of ABA on stomata, a previously
well-watered tomato plant was not watered for five days. The
following measurements were taken:
• the water potential of leaves
• the resistance to the flow of air through the stomata
• the ABA content of leaves
-508 kPa
B
C
-552 kPa
Suggest, with an explanation, the range of possible values for
the water potential of cell B (2).
When the stomata are fully open, the resistance to the flow of air is
low. The plant was watered again at the beginning of day 6 of the
investigation.
2. Cylinders of potato were cut using a cork borer and dried. Each
cylinder was placed into one of a range of sucrose solutions of
different concentrations. The cylinders were left for 8 hours and
then removed from the solutions and redried. The mass of each
cylinder was recorded before and after immersion. The graph
shows the results obtained.
Fig 7. Results of the investigation
0
+10
leaf water
potential
-0.8
inecrease
leaf water
potential
mPa
conc.of
sucrose/
mol dm-3
% change
in mass
-1.6
0.1
0.2 0.3
0.4
0.5
0.6 0.7
0.8 0.9
1.0
decrease
4
80
70
-10
resistance
to air flow
60
resistance
to air flow 5 0
/arbitrary
40
units
3
30
ABA
content
2 of leaves/
arbitrary
units
20
1
ABA content
(i) Explain why the change in mass is expressed as a percentage
change(1).
(ii) Explain fully the shape of the curve (6).
3. Xerophytes are plants that are adapted to grow in soils where
water is scarce. Often the soil water potential is around –50kPa.
Suggest a likely figure for the water potential of the cell contents
of a xerophyte’s roots and explain why this is essential for the
survival of the xerophyte.(2)
10
0
0
1
2
3
4
5
6
days
7
8
9
10
4. Potatoes were cut into two sizes of cubes - 1cm3 (1cm x 1cm x
1cm) and 27cm3 (3cm x 3cm x 3cm). After weighing, twentyseven of the 1cm3 cubes were placed in a beaker of distilled
water. One 27cm3 cube was placed in another beaker of distilled
water. The cubes were left for ninety minutes then removed,
dried and reweighed. The results are shown in the table.
Typical Exam Question 7
Comment on the hypothesis that ABA is synthesised during
times of water stress to cause stomata to close (3).
ABA rises as water potential falls;
Resistance to air flow rises as water potential falls;
Correlation between ABA and air flow resistance;
Correlation does not prove causal link;
3
Volume of cube/cm
Surface area of cube/cm2
No. of cubes in beaker
Surface area:volume ratio of one cube
Total mass of cubes at start/g
Total mass of cubes after 90 minutes
in water/g
Increase in mass/g
Percentage increase in mass
1cm3 cubes
1
6
27
3cm3 cube
27
54
1
52.20
54.48
51.80
52.38
Markscheme
(a) Complete the table (4).
(b) (i) Why were 27 small cubes used in this investigation? (1)
(ii) Comment on the difference in the percentage increase in
mass of the different sized cubes (2)
5
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225. Synoptic Biology: Water Potential
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Markschemes
1. Range from –509 to –551 kPa;
Water moves from A to B, so B must be lower than –508;
Water moves from B to C, so B must be higher than –552;
2. (i) Initial mass of cylinders not idenitical;
To be able to directly compare the results;
(ii) Between 0.0 and 0.34 mol dm-3 water moves into potato cells;
By osmosis;
Because the water potential of the cells is less than that of the solution;
So mass increases;
Concentration of sucrose solution equivalent to the mean water potential of the potato cells is circa 0.35;
Because there is no mass change/no net osmosis;
Below 0.34 mass decreases;
Because water is drawn out by osmosis;
Little/no water remaining in potato/fully plasmolysed/all water has moved out;
Cell wall prevents further shrinkage
3. Any reasonable value less than –50kPa;
to enable the plant’s root hairs to draw water from the soil;
by osmosis/down the water potential gradient;
4. (a)
3
Volume of cube/cm
Surface area of cube/cm2
No. of cubes in beaker
Surface area:volume ratio of one cube
Total mass of cubes at start/g
Total mass of cubes after 90 minutes in water/g
Increase in mass/g
Percentage increase in mass
1cm3 cubes
1
6
27
6.1
52.20
54.48
2.28
4.36
3cm3 cube
27
54
1
2.1
51.80
52.38
0.58
1.11
(b) (i) To give the same volume of potato / only surface area different;
(ii) Smaller cubes increased in mass more than larger cube;
Larger surface area:volume ratio);
Enabling faster osmosis;
Acknowledgements:
This Factsheet was researched and written by Kevin Byrne.
Curriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU.
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ISSN 1351-5136
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