INTRODUCTION
Wate Waterr is absol absolute utely ly esse essenti ntial al for all all living living orga organis nisms ms.. Most Most organ organism isms s are are compris comprised ed of at least least 70% or more more water. water. It provides provides structural structural support support via turgor pressure in plants. Some plants, like a head of lettuce, are made up of nearly 95% water. It is utilized by the cells as a solvent for the uptake and transport of materials such as molecules. The diffusion of water across a semi-permeable membrane is called osmosis. In this process, water moves from an area of higher energy to an area of lower energy. Water potential is the measure of free energy of water in a solution. Water potential (Ψ) is based on Ψ s, Ψp, and Ψm, according to the following formula:
Ψ = Ψs + Ψp + Ψm
Water potential (Ψ) is pressure/unit pressure/unit volume, and is the measure of the tendency tendency of water to move from a system to another and is expressed by MPa. Solute potential, also called osmotic potential, (Ψ s) represents the effects contributed contributed by the dissolved solutes in water; pressure potential (Ψ p) represents represents the effects contributed by pressure, whereas whereas matric matric potentia potentiall (Ψ m) repre represen sents ts the effect effects s contr contribu ibuted ted by water water-b -bin indin ding g colloids in the cells. Potatoes (Solanum tuberosum) are important food crops in the cooler regions of the world. The tuber, the edible part of the white potato is a very short and thick, starchy stem, with the "eyes" being the buds on the stem. White potatoes have firm tissue and convenient size, so they are favorite subjects for the laboratory determination of the water potential of plant tissue.
In this experiment, the group is expected to measure the water potential and osmotic potential in a tissue, and then calculate the pressure potential.
MATERIALS AND METHODS
Potato Tubers
Blender
Sucrose (C 12H22O11) solutions (0.1 m;
Heidenhain (freezing) thermometers
0.2m; 0.3m; 0.4m; 0.5m; 0.6m; 0.7m)
Potato Peelers
Distilled water
Knife
No.5 Cork borer
Ruler
Balances and weighing paper
Cheesecloth
Plastic cups Funnel, 150 mm Paper towels
Part 1: Determination of water potential
Eight different sucrose concentrations (Distilled water, 0.1m, 0.2m, 0.3m, 0.4m, 0.5m, 0.5m, 0.6m, 0.6m, 0.7m) 0.7m) were were prepared prepared and were dispensed dispensed separately separately into plastic plastic cups. cups. Using the cork borer, 16 cylinders were bore from a large potato and were cut into 4 cm slices. The cylinders were blotted with paper towels and were weighed in sets of two. The weight was recorded as the initial weight (W i). One set of cylinders were placed in each of the plastic cups containing the sucrose solutions. After 30 minutes, the cylinders were removed, blotted with paper towels, and were weighed again. The weight was recorded as the final weight (W f ). ). The change in weight (∆W) and the percent change in weight (%∆W) were computed using the following formula: ∆W = Wf - Wi %∆W = ∆W
Wf
All results were tabulated. The % change in weight was plotted against sucrose concentration, the best-fit line drawn through the points. The molal concentration of sucrose that gives 0% change in weight was determined. The Ψ s in bars of that sucrose solution was computed using the following formula: Ψs = -miRT
In this equation, m is the molal concentration of sucrose that gives 0% change in weight, i is the ionization costant (1 for sucrose), R is the gas constant (equal to 8.31 J K-1 mol-1), and T is the room temperature expressed in Kelvins (°C + 273). The water potential of the potato was determined under the assumption that the value of Ψ
m
is
small and therefore negligible.
Part 2: Determination of solute potential Ψ s of extracted sap by cryoscopy
The leftover potatoes from part 1 were chopped and pureed in a blender. The blended potato was filtered using cheesecloth, its filtrate was placed in a beaker.
The temperature of the crushed ice-salt bath was obtained by immersing the heidenhain thermometer. The reading was recorded. 60 mL of the sap was placed in a beaker. The thermometer thermometer was inserted into the beaker and was placed into the ice-salt bath, stirring vigorously. vigorously. When the temperature temperature read 1°C, the the temperature was was read and recorded every 10 seconds. A plot of the temperature versus time was made.
RESULTS AND DISCUSSION
In the cups containi containing ng distille distilled d water water 0.1, and 0.2 molal sucrose sucrose solution solutions, s, the potatoes increased in weight because water moved in due to the potato’s lower water potential with respect to the solution. Hence, the percent change in mass is positive.
Sucrose Concentration 0 (Dist.H20) 0.1 m 0.2 m 0.3 m 0.4 m 0.5 m 0.6 m 0.7 m
Initial Weight (in grams) 3.3660 3.3493 3.7560 3.4850 3.8923 3.9682 3.5191 3.5153
Final Weight (in grams) 3.5856 3.5185 3.8711 3.4440 3.6653 3.6555 3.0111 2.9189
Change in Weight 0.2196 0.1692 0.1151 -0.0410 -0.2270 -0.3127 -0.5080 -0.5964
% Change in Weight 6.12 4.81 2.97 -1.19 -6.19 -8.55 -16.87 -20.43
In the cups containing 0.3 - 0.7 molal sucrose solution, the potatoes decreased in weight because water moved out due to the potato’s higher water potential with respect to the solution. Hence the percent change in mass is negative.
Figure 1 Percent Change in Weight vs. Sucrose Concentration
The graph depicting the percent weight change (y) as a function of the sucrose concentration (x) is shown in Figure 1. Because none of the given solutions gave a 0% change in weight, the equation of the best-fit line (Figure 1) was used to solve for the exact molal concentration needed to produce the desired effect. Hence,
The equation of the best-fit line:
y = -39.494x + 8.9892
If percent weight change is 0:
0 = -39.494x + 8.9892
-8.9892 = -39.494x x = -8.9892-39.494 x = 0.22760 ≈ 0.23
Hence, Hence, if a potato potato cylinder cylinder is immersed immersed in a 0.23 molal sucrose sucrose solution, solution, the potatoes will retain the same weight because the water moving out will be balanced by the water moving into the potato cells since the solution’s water potential is equal to the potatoes’ water potential.
Knowing Ψ p=0MPa in a free standing solution and Ψ m is neglected, then Ψ = Ψs, the water potential of the potato can be solved by the following equation:
Ψs = -miRT =(-0.23 molal x 1 x 10 3 mol m-3) (1) (8.31 J K -1 mol-1) (25 + 273K) = -569 567.4 J/m 3 / 106 Ψs = - 0.57 MPa
Solute potential (Ψ s) represents the effect of solutes on the energy state of water. Solute potential is related to other properties of the solution such as vapor pressure, boiling boiling point, point, and freezing freezing point. These These properti properties, es, which which are interrel interrelated ated,, are called called colligative properties, and are dependent on the mole fraction of solute. Since these properties are inter-related, one can be measured and used to calculate the others. The cryosco cryoscopic pic osmomet osmometer er measure measures s the osmotic osmotic potential potential of a solution solution by measur measuring ing its freezing freezing point. point. Solution Solutions s have have colliga colligative tive properti properties es that collecti collectivel vely y depend on the number of dissolved particles and not on the nature of the solute. One of the colligative colligative properties of solutions is the decrease in the freezing point as the solute concentration increases. A 1 molal solution of an ideal non-ionized solute has an solute potential of -2.27 MPa and freezes freezes at -1.86 -1.86 C. Based Based on this relation relationship ship the solute solute potential potential of any unknown solution can be calculated: Ψs = (1.22 MPa deg -1 ) Tf,
where Tf is the freezing point of the solution in oC.
Since this equation is for solutions at zero oC (273 K) the equation must be corrected to obtain the answer at room temperature by multiplying the equation by the correction factor which is the ratio of the absolute temperatures (room temperature in K/273 K). All in all: Ψs = (1.22 MPa deg -1 ) Tf x (room temp in K/273 K)
Figure 2 Temperature versus time graph of the extracted sap
In a typical freezing point curve, there is a continuous drop in temperature as the liquid is cooled, then it remains constant constant signifying that both liquid and solid phases are present. present. Temperature finally drops again when only solid phase is present. From the tempera temperatur ture e vs. time curve shown shown above, above, the temper temperatur ature e became became stable stable for thirty thirty secon seconds ds at -3.6° -3.6°C. C. This This will will be the the desig designat nated ed freez freezing ing point point for for the the sap sap extra extract. ct. However, this is merely an estimate and is called the apparent freezing point. The true freez freezing ing point point (Tf ) is obta obtaine ined d after after correc correctin ting g for supe super-c r-coo oolin ling g accor accordin ding g to the the following equation:
Tf = Tf ' - 0.0125 t s ts = Lowest temperature - T f ' Where Tf is the true freezing point and T f ' is the apparent freezing point. t s is the degrees of supercooling (negative in sign), while 0.0125 corresponds to the amount of water (1/80) that solidifies per degree of supercooling. Tf = Tf ' - 0.0125 t s = -3.6 – 0.0125 (-3.7 - -3.6) = -3.6 – (-0.00125) = -3.59875 ≈ -3.6 Correction also has to be made for the zero point of the thermometer to determine the solute potential. Tf = Tf ' - 0.0125 t s = 0 – 0.0125 (-3.7 – 0) = -0.0463 ≈ -0.05 The solute potential at 0°C can now be determined: Ψs = (1.22 MPa deg -1) Tf x (temp in K/273 K) = (1.22) (-3.6) °C x [(0+273)/273 K] = -4.392 MPa At room temperature however, Ψs = (1.22 MPa deg -1) Tf x (temp in K/273 K) = (1.22) (-3.6) °C x [(25+273)/273 K] = -4.794 MPa
Lastly, the pressure potential (Ψ p) of the cells of the potato can be solved by looking back at the formula for water potential: Ψ = Ψs + Ψp (Ψm is neglected) Under equilibrium conditions, the water potential of the potato is equal to the water potential of the 0.23 molal sucrose solution. Therefore, the water potential (Ψ) of the potato is -0.57 MPa. The computed solute potential of the potato sap extract at room temperature is -4.794 MPa.
Ψ = Ψs + Ψp -0.57 MPa = -4.794 MPa + Ψ p Ψp = 4.794 MPa - -0.57 MPa Ψp = 4.224 MPa
REFERENCES
Journals: Bland, W. L. and C. P. Tanner. 1985. Measurement of the water potential of stored potato-tubers. Plant Physiology 79: 891-895. Boyer, J. S. 1969. Water status measurements in plants. Annual Reviews of Plant Physiology 20: 351-364. Books: Meyer, B. S. and D. B. Anderson. 1935. Laboratory Plant Physiology. Edwards Bros., Ann Arbor. 107 pp. Salisbury, F. B. and C. W. Ross. 1992. Plant Physiology , 4th Ed. Wadsworth Publishing Co., Belmont, CA. 682 pp. Website: Saupe, S. G. (2009, January 7). Determining Osmotic Potential by the Freezing Point Depression Method . Retrieved November 22, 2010, from Plant Physiology (Biology 327) Home Page: http://employees.csbsju.edu/SSAUPE/biol327/Lab/water/water-lab-freez.htm
EXPERIMENT 1 CELL WATER POTENTIAL
SUBMITTED BY: GROUP No. 8 Buenaflor, Maria Katrina A. De Leon, Eugene Morada, Jayvee V. Sison, Marcus Isaiah
SUBMITTED TO: Mr. Josefino Castillo
DATE SUBMITTED: November 25, 2010
