Biology 3460 Week 5 1 Biology 3460 - Plant Physiology - Lab Exercise 5 Plant Water Relations: Pressure Chamber and Osmosis Objectives: This lab is intended to: (1) investigate the consequences of water stress on the water potential in bean plants, (2) gain practical experience with detering leaf xylem water potential using the pressure bomb, (3) provide experience diluting solutions to make known solution concentrations, (4) provide examples of how solute concentration can influence water potential, and (5) permit calculation of osmotic potential using the Van't Hoff equation. Introduction Plants do NOT use active transport to move water into cells. In plants, water moves across the selectively permeable cell membrane by a passive process controlled by gradients in water potential (similar to chemical potential). The term water potential ( or w ) is used to describe the sum of forces involved in water movement, and water potential is typically expressed in SI units of mega pascals (MPa). Other non-si units are also used to measure water potential; atmospheres (atm) or bars (bar). Since water loses energy as it moves passively, it must move from an area of high water potential to an area of low water potential. If the water potential on each side of the membrane is equal, there is no net movement of water across the membrane. When this happens, we say that equilibrium has been reached; the water potential inside and outside the cell is the same. Three factors (pressure potential, solute potential, and gravity potential) contribute to w but with respect to cellular water relations, we will simply sum the solute potential ( s ) and pressure potential ( p ) to give us total cell water potential ( w ). The gravity term is a or effect that only becomes significant when dealing with water movement to the tops of very tall trees. Mathematically, water potential within cells can be expressed as: w = s + p The following exercises will emphasize both water potential and solute potential, and the effects of water stress on water potential. Plant water stress occurs when the evaporative demand exceeds the water supply and can have a significant impact on photosynthesis. Generally, the immediate response to water stress is an increase in resistance due to stomatal closure and this restricts the movement, or flux, of carbon dioxide and water vapor. The balance between water and carbon dioxide flux in a leaf is crucial in the photosynthetic process. In this lab, you will use a pressure bomb to detere the xylem water potential in unstressed and water stressed bean plants. Later, the effect of various environmental factors on photosynthesis will be reinforced using computer simulations.
Biology 3460 Week 5 2 Part A. Deteration of Leaf Xylem Water Potential using a Pressure Bomb The water potential of a leaf at equilibrium is similar throughout the leaf. However, the contributions of various components typically vary in different cells and tissues. In the xylem, the major contribution is a negative pressure potential. Generally xylem is thought of as containing pure water (with very little dissolved solutes). In this manner, we can consider the xylem pressure potential to be equal to its water potential. We know that w = s + p. However, in xylem s ~ 0, thus w = p. When a leaf is excised from a plant, the tension in the xylem is released ( p = 0). As the water moves along a water potential gradient into surrounding cells, the water recedes from the cut surface. We can detere the amount of pressure required to force the water back to the cut surface by placing the leaf in a pressure bomb and applying pressure. The amount of pressure is equal but opposite in sign to the tension (negative pressure) that existed in the xylem at the time when the leaf was cut (i.e.: applied pressure = - p ). Protocol: Work as a group of three students to complete this part of the lab. 1. Choose a Phaseolus plant from each of the two treatments: unstressed and waterstressed. When selecting your plants, choose plants that have three to four fully expanded leaves. 2. Complete the following for each of your plant samples. It is important not to mix leaf tissue from different treatments! Use a razor blade to cut the youngest mature leaf at the petiole (cuts should be perpendicular to the axis of the petiole). Your leaf sample should have as much petiole attached as possible (i.e. make your petiole cut near the stem). 3. Carefully insert the petiole into the widest end of the rubber stopper using the insertion tool. Your instructor will demonstrate this. The petiole should fit snugly into the stopper and protrude from the stopper end (Figure 1). 4. Ensure that the control valve is in the EXHAUST position. Firmly press the stopper into the hole in the chamber lid. Place the lid on top of the chamber, pressed down, and rotate the lid clockwise to close. Be familiar with the pressure tank, connections, and valves before operating the instrument - it can be a BOMB and cause serious damage!!
Biology 3460 Week 5 3 Cut end of petiole Rubber bushing (stopper) Thick steel walls of pressure bomb chamber Pressure gauge Pressure from cylinder of nitrogen Figure 1. The pressure bomb. 5. You will NOT need to adjust the valve on the nitrogen cylinder; this will be done for you. Do NOT place any part of your body directly above the chamber opening. Turn the control valve to the CHAMBER position. Adjust the rate valve as necessary to increase the pressure in the chamber slowly (about 0.3 bars/s) while observing the cut end of the petiole through the magnifying glass. When the cut surface is just wetted, turn off the pressure by rotating the control valve to the OFF position, read the gauge, and record the value in Table 1. Do NOT use the rate valve to stop the pressure. NOTE: phloem sap may exude around the outside of the petiole even under a low pressure. If many bubbles are present, you have missed the endpoint. 6. Turn the control valve to the EXHAUST position, remove the chamber cover by rotating it counter-clockwise, and save the leaf tissue (place in marked envelopes).
Biology 3460 Week 5 4 Table 1. Water potential values for unstressed and drought-stressed Phaseolus vulgaris plants. Plant Condition Unstressed Drought Stressed Water Potential (MPa*) Plant 1 Plant 2 Plant 3 Plant 4 Plant 5 Mean SD * Most pressure bombs are calibrated in bars, to convert from bars to MPa divide by 10. Analysis: 1. Use class results to prepare a figure that shows the effect of water stress on the water potential of bean leaves. Mean water potentials with standard deviations should be plotted. 2. Using an appropriate statistical test, detere if there is a significant difference between the treatments. 3. What effect did the water stress treatments have on water potential? Why? Explain. Part B. Deteration of Water Potential of Potato Tuber Segments A cell membrane is largely impermeable to large molecules such as sucrose. However, smaller molecules such as water can freely move across the membrane. Plant tissue placed in a hyperosmotic solution (the solute concentration is greater in the solution compared to the cellular interior) will lose water to the solution. Plant cells placed in a hypoosmotic solution (the solute concentration is less in the solution relative to inside the cell) will gain water from the solution. Tissue placed in an isoosmotic solution will neither gain nor lose water since the solute concentration of the external solution is the same as the internal cytoplasm. Depending on the relative water potentials, water will also move from a less negative (higher) to a more negative (lower). In this exercise, you will detect water movement by measuring the weight of potato tissue before and after incubation in a series of sucrose concentrations. The sucrose concentration that results in no change in weight may be assumed to have the water potential equal to that of the tissue. Using the Van't Hoff equation, you will calculate the water potential of the potato tissue. Protocol: Work as a group of three students to complete this part of the lab. 1. Label seven 100-mL beakers for the sucrose solutions listed: 0, 0.1, 0.2, 0.3, 0.4, 0.5, and 0.6 M. Calculate the volume of 1.0 M sucrose solution and distilled water needed
Biology 3460 Week 5 5 to make 50 ml of each of the above solutions. Enter these values in Table 1 in the appropriate space. Check with your instructor to make sure your volumes are correct before proceeding. 2. Using the 1M stock sucrose solution and distilled water create 50 ml of each of the concentrations listed above in the labeled beakers. 3. Working quickly but carefully, use a cork borer to obtain 21 cylinders from potato tuber tissue. To remove the tissue from the borer, use the dowel provided. Trim the cylinders to ~3 cm in length. Place the prepared potato segments in a covered Petri plate to prevent desiccation. 4. Blot each cylinder with a paper towel, weigh groups of three (3) segments to the nearest 0.01 g, and record the weight in Table 1. Place one set of three segments in each prepared sucrose solution and note the time. Be sure to keep track of which cylinders are going into each solution!! 5. After a 20 ute incubation period, remove each group of three segments, blot them on a clean, dry, paper towel and reweigh them to the nearest 0.01 g. Work quickly and carefully. Record the weight in the proper place in Table 1. Immediately return the segments to their respective solutions and continue tig. Table 1. Effect of various sucrose concentrations on water uptake in potato segments. Volume of 1M sucrose (ml) Volume of water (ml) Final sucrose concentration (M) 0 20 Weight of cylinders (g) 40 60 80 100 0.0 0.1 0.2 0.3 0.4 0.5 0.6
0 1 2 3 4 5 6 Biology 3460 Week 5 6 6. Repeat step 4 until there is no net change in ANY of the groups of segments. Which of the cylinders seem crispy? Which of the cylinders appear limp? 7. Discard the potato cylinders and their respective sucrose solutions. Place the glassware on trays provided on the side bench. Analysis: 1. Subtract the initial segment weights from the final weights to find the change in weight for each sucrose treatment. Enter these values in Table 2 for the appropriate sucrose concentrations. Divide the difference by the initial weight and multiply by 100 to calculate the percent weight change. Enter this value in the appropriate space in Table 2. Table 2. Change in weight and percent change in weight of potato segments incubated in various sucrose concentrations. [Sucrose] (M) 0 0.1 0.2 0.3 0.4 0.5 0.6 Change in Weight (g)* % Change in Weight** * Change in weight = Final weight - Initial weight ** % Change in weight = (Change in weight / Initial weight) 100 (retain sign to indicate gain or loss) 2. Plot the percent change in weight against the various sucrose concentrations on the graph provided. 600 500 400 300 200 100 0
Biology 3460 Week 5 7 3. Detere the sucrose concentration that gives 0% change in weight. In a freestanding solution (open solution), there is no turgor pressure ( p = 0), so the of the solution is equal to the s of the solution (i.e.: = s ). Calculate the s of that sucrose solution using the formula below, the Van't Hoff equation. Note that C must be converted from mol L -1 to mol m -3 in order for unit cancellation. Your answer will be in the units of J m -3. To convert your answer to MPa, divide by 10 6. s = - CiRT where: C is the molar concentration of the solutes (molarity = moles L -1 ; 1 molar = 1 X 10 3 mol m -3 H 2 O), i is the osmotic coefficient (the value of i is 1 for molecules that do not dissociate in solution and can be 2 or more for molecules that completely dissociate, i.e. salts), R is the gas constant (8.31 J K -1 mol -1 ), and T is the absolute temperature (room temperature in K, i.e.: C + 273). What is the water potential of the potato tissue?