5_experiment 1_cell Water Potential.pdf

COLLEGE OF SCIENCE DEPARTMENT OF BIOLOGICAL SCIENCES Experiment 1 CELL WATER POTENTIAL

The tendency of water to move into or within a system, such as a plant or animal tissue, the soil or the atmosphere can be measured as the amount of energy per unit volume (or pressure). Such values, expressed in units of bars or megapascals (MPa) is referred to as the water potential of the cell and may be computed mathematically as: Equation (1 –1): where:

Ψ = Ψs + Ψ p + Ψ m

Ψ = water potential of a cell; Ψ = 0 at equilibrium; Ψs = solute or osmotic potential; always negative in sign; Ψp = pressure potential; turgor pressure; may be (-), 0, or (+); Ψm = matric potential; due to water-binding colloids in the cell; negative in sign.

Plant cells usually have a zero (0) water potential when they are in equilibrium with pure water (turgid). Dissolved solutes contribute to the water potential. The objective of this experiment is to measure the water potential and osmotic potential in a tissue, then calculate the pressure potential. For practical reasons, a plant tissue, instead of animal tissue, is used in this experiment. Materials: Plant samples (potato, sweet potato, radish, turnips, cucumber)* Sugar solutions (0.1m; 0.2m; 0.3m; 0.4m; 0.5m; 0.6m; 0.7m) Distilled water Balances and weighing paper Plastic cups Basin Sugar Salt Funnel, 150mm Erlenmeyer flasks, 250 ml Paper towels/filter paper Blender Heidenhain (freezing) thermometer / Radioshack Infrared Thermometer Chopping board Magnetic stirrers with stir bars Parafilm Knife Ruler Cheesecloth *Laboratory instructor may modify this list. Do not wash the plant samples. 1|P a g e

COLLEGE OF SCIENCE DEPARTMENT OF BIOLOGICAL SCIENCES PART I. DETERMINATION OF WATER POTENTIAL, Ψ When a tissue is placed in a sugar solution, there will be a net movement of water into or out of the tissue depending on the relative water potentials: into the tissue with a hypotonic solution; out of the tissue in a hypertonic solution; or no net movement in an isotonic solution when the tissue and solution are at equilibrium. If a series of solutions of different concentrations of sucrose is used, the sucrose concentration that would cause no change in weight may be determined by plotting the percent change in weight against sucrose concentration. Such solution may be assumed to have a water potential equal to that of the tissue. Procedure: 1. Prepare eight (8) different sucrose concentrations (dH2O, 0.1m, 0.2m, 0.3m, 0.4m, 0.5m, 0.6m, 0.7m) and dispense 75ml of each separately into 150ml beakers / plastic cups. 2. Cut plant samples into small pieces approximately 1 cubic centimeters each. Place 3 – 4 cubes in covered beakers to prevent them from drying out. Save remaining plant samples for Part II. 3. Blot cylinders with paper towels and weigh (to 0.001g) in sets of 3 - 4. Record the weight. 4. Put one set of cubes in each of the beakers with sucrose solutions prepared in Step 1. 5. After 45 minutes to 1.0hr, remove the cubes, blot with paper towels and weigh again. Record results as before. 6. Compute for the change in weight (∆W) and the percent change in weight (%∆W) using the following formulae. Equation (1–2):

∆W = W f − Wi

Equation (1–3):

% ∆W =

where:

∆W x100 Wi

Wi = initial weight of the cylinders; Wf = final weight of the cylinders.

7. Tabulate all results. Plot % change in weight vs. sucrose concentration. Draw the best-fit straight line through the points. 8. Determine molal concentration of sucrose that gives 0% change in weight. Compute for Ψs in bars of that sucrose solution using the formula: Equation (1–4): where:

Ψs = -miRT

m = molality; (NOTE: 1 molal = 1 x 103 mol m-3 H2O) i = ionization constant = 1 for sucrose R = gas constant = 8.31 J K-1 mol-1 T = room temperature in K (oC + 273)

NOTE: For uniformity, room temperature is computed at 26oC 2|P a g e

COLLEGE OF SCIENCE DEPARTMENT OF BIOLOGICAL SCIENCES 9.

Determine the water potential of the potato assuming that Ψm is very small and therefore negligible. Convert unit of J m-3 (energy per unit volume, pressure which is equivalent to Pa) to MPa by dividing computed value by 106. NOTE: In a free standing solution, Ψp = 0; so Ψ = Ψs.

PART II. DETERMINATION OF SOLUTE POTENTIAL (ΨS) OF EXTRACTED SAP BY CRYOSCOPY Cryoscopy (cry·os·co·py) is the examination of liquids, based on the principle that the freezing point of solutions varies according to the amount and the nature of the solute. This provides a relatively easy means of arriving at the osmotic potential of a solution. The freezing point of an aqueous solution decreases with an increase in dissolved solutes. Once the freezing point of a solution has been determined, its solute potential may be calculated, thus: Equation (1–5): where:

Ψs = 1.22Tƒ Ψs = solute potential in MPa at 0oC Tƒ = freezing point of the solution in oC

The equation holds true for freezing temperature. The solute potential of the extracted sap at room temperature can then be computed: Equation (1–6): where:

Ψs = 1.22Tƒc c is the correction factor; (Rm temp in K/273K)

Procedure: Sap extraction 1. Peel 2 – 4 potatoes, chop and place in a blender. Leftovers from Part I may be used. Puree the tissue. 2. Filter blended potato with cheesecloth to remove cell wall and debris. Store in covered beaker / plastic cup. Freezing point determination 3. Immerse Heidenhain thermometer in crushed ice – salt bath. Record reading. 4. Pour 60ml sap in 250ml Erlenmeyer flask with magnetic stirring bar and insert thermometer. Surround flask with ice–salt bath (see Figure 1–1). Stir vigorously. 5. When temperature reads about 1oC, read temperature every 10 sec. NOTE: There should be a continuous drop in temperature, followed by a short plateau which corresponds to the freezing point, is followed by an increase. 6. Plot temperature vs. time.

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COLLEGE OF SCIENCE DEPARTMENT OF BIOLOGICAL SCIENCES

Heidenhain thermometer Erlenmeyer flask

Rubber stopper

Beaker Ice-salt bath

Stir bar

Sap extract

Magnetic stirrer

5 4

6

7

3 2 1

5 4 8

3

9

2

11

6

7 8 9

1

10

Figure 1 – 1. Setup for freezing point determination 7.

Determine the true freezing point, Tƒ. Equation (1–7): where:

Tƒ = Tƒ’ – 0.0125ts

Tƒ’ = apparent freezing point ts = degrees of supercooling; (-) in sign ts = lowest temp - Tƒ’ 0.0125 = amount of water (1/80) solidifying per degree of supercooling

8.

Make another correction for the zero point of the thermometer.

9.

Determine the solute potential using Equation (1–5) at 0oC.

10.

Calculate the pressure potential (Ψp) of the cells of plant sample.

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COLLEGE OF SCIENCE DEPARTMENT OF BIOLOGICAL SCIENCES Group No. ____ Group Members: ___________________________ ___________________________ ___________________________ ___________________________ ___________________________

Date Performed: ____________________ Lab Instructor(s):____________________ ______________________ Signature(s): ______________________

Experiment 1 CELL WATER POTENTIAL I. Determination of Water Potential Sucrose Concentration

Initial Weight In Grams (Wi)

Final Weight In Grams (Wf)

Change in Weight (∆W = Wf - Wi)

% Change in Weight (%ΔW)

Dist. H2O 0.1 m 0.2 m 0.3 m 0.4 m 0.5 m 0.6 m 0.7 m Table 1-1. Results for determination of water potential. Molality of sucrose where ΔW = 0 Show your computations below:

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COLLEGE OF SCIENCE DEPARTMENT OF BIOLOGICAL SCIENCES II. Determination of Solute Potential of extracted sap through cryoscopy Temperature (Celsius)

Time (seconds)

Temperature (Celsius)

Time (seconds)

Table 1-2. Results for determination of solute potential. III.

Compute for the water potential of your plant sample. Solute Potential (Ψs)

IV.

Pressure Potential (Ψp)

Water Potential (Ψ)

Summary of Results

Group No.

Plant Sample (Common Name, Scientific Name)

Ψ

1 2 3 4 5 6 7 8 9 10 Note: Water potential should be indicated up to three (3) decimal places.

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COLLEGE OF SCIENCE DEPARTMENT OF BIOLOGICAL SCIENCES

Guide Questions: 1. Why is the osmotic potential always negative in sign? ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ 2. How do dissolved solutes contribute to the water potential of the cell? ______________________________________________________________________ ______________________________________________________________________ _____________________________________________________________________ ______________________________________________________________________ 3. What is the sign of the pressure potential of the tissues at equilibrium? (+), (0), or (-)? At incipient plasmolysis? During transpiration? ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ 4. What causes supercooling? Why is the lowest point reached not considered the freezing point of the sap solution? ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ 5. Do plants with watery sap have higher or lower water potential? Explain your answer using actual computed values. ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________

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