Synoptic Biology: Water Potential

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1 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 Fig. shows the water potential of four plant root cells. 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 Ψ = -85 Ψ = -9 Ψ = -5 Cell A Ψ = -7 kpa Cell B Ψ = -3 kpa Ψ = -3 Cell C Ψ = -975 kpa Cell D Ψ = -2 kpa Each cell was placed into a sucrose solution with a water potential of -54 kpa. All of the cells took in water from this solution. (a) Which cell would initially take in water at the fastest rate? Explain why (3). (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! Draw arrows on Fig.2 to show the net flow of water between the cells (3). Water will always move towards the most negative number. So hopefully, you drew arrows as follows: arrow drawn from -85 to -9, from -85 to -5 and from -5 to -3 Exam Hint:- Did you draw a continuous arrow from -85 to - 5 to 3? 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. (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; The next exam question takes this one small step further by considering how water potential affects water movement inside the plant. Extract from Chief Examiner s 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 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.

Fig 3 shows the appearance of a plant cell when immersed in three different concentrations of sugar. Fig 3 cell wall A B C cytoplasm vacuole The cell has absorbed water from the solution.

The water potential of the solution must be less than that of the cell. Almost all of the water has been osmotically drawn out of the cell vacuole.

Molarity of sucrose solution/mol dm -3..2.3.4.5 Initial mass/g.2.7.25.9.8.7 Final mass/g.5.32.3..3.93 % change in mass 3. 2.8 4. - 7.3-2.7-2.

2 Fig 3 shows the appearance of a plant cell when immersed in three different concentrations of sugar. Fig 3 cell wall A B C cytoplasm vacuole The cell has absorbed water from the solution. The water potential of the cell must be lower than that of the surrounding solution. Water has been osmotically drawn out of the cell. The water potential of the solution must be less than that of the cell. 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. Table shows the results of such an investigation. Molarity of sucrose solution/mol dm Initial mass/g Final mass/g % change in mass 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. 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 2 hours Extract from Chief Examiner s Some candidates continue to confuse plasmolysis and turgidity. % change in length.2.4. concentration of sucrose /mol dm -3 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. 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:. 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 5. Add the cylinders to a range of sucrose solutions of known molarity. 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. / 2 -.5/G (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. mol dm -3 (2). (ii) Suggest why there was no change in mass of the potato placed in a sucrose solution of.2mol dm -3. (iii)explain the results obtained when the potato cylinders were placed in sucrose solutions between.4 -.mol 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) (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;

Surface area: Volume ratios The effect of the surface area: volume ratio on osmosis can be investigated using cubes of potato. Typical Exam Question 4 Cubes were cut from fresh potato.

3 Surface area: Volume ratios The effect of the surface area: volume ratio on osmosis can be investigated using cubes of potato. Typical Exam Question 4 Cubes were cut from fresh potato. The dimensions of the cubes are shown in the diagram cm x cm x cm Surface area = cm 2 Volume = cm 3 2cm x 2cm x 2cm Surface area = 24cm 2 Volume = 8cm 3 cm 2cm cm cm 2cm 2cm The cubes were blotted dry with filter paper and weighed. One cube measuring 2cm x 2cm x 2cm was placed in a beaker and covered with distilled water 8 cubes, each measuring cm x cm x cm were put into a second beaker and covered with distilled water At intervals, for a period of 4 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. % increase in mass time/hours 8 cubes of side cm x cm x cm cube of side 2cm x 2cm x 2cm (a) Suggest why 8 cubes of dimension cm x cm x cm were used in the investigation (). (b) Explain the results shown (2) 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 cm cubes was faster than that of the 2cm cube because greater SA:V ratio/ : versus 3:; 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; 3 Xerophytes 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. Extract from Chief Examiner s Candidates need to know that transpiration should be considered in terms of loss of water vapour down water potential gradients Water movement across the root Fig.4 shows the three routes by which water moves across the cells in the route of a dicotyledonous plant. Fig 4 root hair Path vacuolar pathway Path 2 via cytoplasm (symplast) pathway Path 3 via cell walls (apoplast) pathway Common pathways xylem 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. Typical Exam Question 5 Scientists measured the water potential of soil in a field in which barley was being successfully grown. The water potential was 42kPa. 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? -38kPa -42kPa -58kPa kpa 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. 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.

Seawater flooding 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.

4 Seawater flooding 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 Potometers (Fig, ) 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. Potometer Typical Exam Question Explain why, after flooding with seawater, plants may wilt rapidly (2) leafy shoot reservoir Capillary tube filled with water 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; water air bubble Mass flow in the phloem 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. Fig. 5 Physical model to illustrate the principle of mass flow tube water Scale 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. Typical Exam Question. The table shows typical values of the water potential at various points between the soil and the top of a tree. Source Sink Location Water potential values (MPa) Soil -.4 Root cells -.2 Stem -.5 Leaf cells -.5 Atmosphere -98 Sugar solution rigid, partially permeable membranes water How it works:. 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. As a mass flow Extract from Chief Examiner s 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. (a) Use the information in the table to outline how water may be drawn from the soil to the top of a tree. (3) (b) State 3 environmental factors which affect the rate of transpiration. (3) Semicolons indicate marking points. (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 4

5 Stomatal opening and closing 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. 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 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 of the investigation. Fig 7. Results of the investigation resistance to air flow /arbitrary units -.8 leaf water potential mpa ABA content resistance to air flow days leaf water potential ABA content of leaves/ arbitrary units Typical Exam Question 7 Comment on the hypothesis that ABA is synthesised during times of water stress to cause stomata to close (3). Practice Questions. 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 -58 kpa C -552 kpa Suggest, with an explanation, the range of possible values for the water potential of cell B (2). 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. + inecrease % change in mass decrease B conc.of sucrose/ mol dm -3 (i) Explain why the change in mass is expressed as a percentage change(). (ii) Explain fully the shape of the curve (). 3. Xerophytes are plants that are adapted to grow in soils where water is scarce. Often the soil water potential is around 5kPa. 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) 4. Potatoes were cut into two sizes of cubes - cm 3 (cm x cm x cm) and cm 3 (3cm x 3cm x 3cm). After weighing, twentyseven of the cm 3 cubes were placed in a beaker of distilled water. One cm 3 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.. 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; Volume of cube/cm 3 Surface area of cube/cm 2 No. of cubes in beaker Surface area:volume ratio of one cube Total mass of cubes at start/g Total mass of cubes after 9 minutes in water/g Increase in mass/g Percentage increase in mass cm 3 cubes cm 3 cube (a) Complete the table (4). (b) (i) Why were small cubes used in this investigation? () (ii) Comment on the difference in the percentage increase in mass of the different sized cubes (2) 5

6 s. Range from 59 to 55 kpa; Water moves from A to B, so B must be lower than 58; 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. and.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.35; Because there is no mass change/no net osmosis; Below.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 5kPa; to enable the plant s root hairs to draw water from the soil; by osmosis/down the water potential gradient; 4. (a) Volume of cube/cm 3 Surface area of cube/cm 2 No. of cubes in beaker Surface area:volume ratio of one cube Total mass of cubes at start/g Total mass of cubes after 9 minutes in water/g Increase in mass/g Percentage increase in mass cm 3 cubes cm 3 cube (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, 5 King Street, Wellington, Shropshire, TF NU. s may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 35-53

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