Physio Lab Experiment 1_cell Potential

Experiment 1: Cell Water Potential Christine Marie H. Ulbis, Adrionna Faye Pauline A. Uy, Gian L. Veloso, Bianca T. Villagomez Department of Biological Sciences College of Science, University of Santo Tomas España, Manila Philippines

Abstract The water potential was measured for each root crops. It resulted to values, -0.7220, -0.7059, 1.0999, -0.8191, -1.2148, -1.2271, -1.3608, -1.3779, -0.6693, -1.0121 from groups 1-10, respectively. Similarly, the osmotic potential for each root crop was also measured, the values for camote ranges from 0.6547 MPa to 3.4722 MPa, for potato it ranges from 0.1325 MPa to 0.4695 MPa, for carrot it ranges from 1.1002 MPa to 1.7801 MPa, for Jicama it ranges from 2.0447 MPa to 3.8304 MPa and for Radish it ranges from 0.4294 MPa to 1.0089 MPa..

Keywords: Osmotic Potential, Water Potential, Pressure Potential, Freezing Point, Root Crops

Introduction

Methodology

Cell Water Potential, symbolized by ψ (the Greek letter psi), is defined as the energy required for water to move from the sample and into a relatively pure free water in reference conditions. Through the process of osmosis, water moves from a region of high area of concentration to a region of low area of concentration. Additionally, concentration, pressure and gravity are major contributing factors in a calculating a plant’s water potential (Sinauer, 2015).

Eight different sucrose concentrations (dH2O, 01.m, 0.2m, 0.3m, 0.4m, 0.5m, 0.6m, 0.7m) were prepared. 75mL of each were dispensed separately into plastic cups. Sixteen cylinders were bore from the root crops (different plants were assigned on each group) using a No. 5 cork borer. Cylinders were placed in covered plastic cups to prevent them from drying out. Each cylinder was cut approximately 4cm and the remaining plant samples were saved for Part B of the experiment. The cylinders were blotted with paper towels and weighed in sets of two. The weights were recorded. One set of cylinders was put in each of the beakers with the said sucrose concentration solutions. After 1.5 hours, the cylinders were removed, blotted with paper towels and weighed again. The weights were recorded. The change in weight and the percent change in weight were then computed. The results were tabulated and the percent change in weight versus sucrose concentration was plotted with the best-fit straight line drawn through the points. The molal concentration of sucrose that gives 0% change in weight was then determined. The water potential of the root crops were the determined.

Cryoscopy is the determination of the depression in freezing point unto which molecular weights of the dissolved solutes can be determined. The freezing point occurs when the temperature of the liquid state of a liquid substance is in equilibrium to its solid state. It has a colligative property meaning the freezing point lowers when a compound or more dissolved solutes are added. Once the freezing point is determined, it can be used to calculate for the solute potential which is part of the equation in calculating for the plant’s water potential. This experiment aims to measure the water potential and osmotic potential in a plant tissue and then calculate the pressure potential.

Based on the computed values for Parts A and B, the water potential of the root crops were determined. The values for all plant samples were then compared.

Results and Discussion In theory, no net gain or loss of water in tissue after immersion in a solution of known molarity means that its water potential is equal to that of the external solution. Employing the Gravimetric Technique, water potential values may be negative due to the infiltration of water and solutes within the apoplast (Saupe, 2009). Bland and Tanner (1985) reported that the osmotic potential for a potato tube is in the range of -0.58 to -1.53 MPa. The root crops should fall within that range. In this experiment, the samples were immersed in solution with a variation of sucrose concentration resulting to the losing or gaining of water by the process of osmosis. Osmosis is the spontaneous net movement of solvent molecules through a semi-permeable membrane into a region of higher solute concentration (Haynie, 2001). If the samples were placed in a solution of higher solute concentration, (hypertonic solution) the water will go out of cell, thus the decrease in the weight of the samples. In contrast, if the samples were placed in a solution of lower solute concentration, (hypotonic solution) the water will go inside the cell, thus the increase in the weight of the samples. This is because dissolved solutes

contribute to the osmotic pressure or potential. Solutes reduce water potential by consuming some of the potential energy available in the water. According to Kramer & Myers (2012), osmotic pressure is the external pressure required to be applied so that there is no net movement of solvent across the membrane (isotonic). It is a colligative property which depends upon the concentration of solute molecules or ions, but not upon the identity of the solute. In these calculations, molality (m) is used which is defined as the number of moles dissolved per kilogram of solvent. This is because the mass of solvent doesn't change with temperature unlike the volume. This is the reason why molality is preferred over molarity (moles per liter). Osmotic pressure and potential, in turn, affects water potential. Water potential is the tendency of the water to move into or within a system measured as the amount of energy per unit volume expressed in megapascals (MPa). Water potential can be computed mathematically as: 𝛹 = 𝛹𝑠 + 𝛹𝑝 + 𝛹𝑚 where 𝛹 is the water potential of a cell, 𝛹𝑠 is the solute or osmotic potential, 𝛹𝑝 is the pressure potential and 𝛹𝑚 is the matric potential. The pressure potential (ψp) is the effect of hydrostatic pressure on the potential energy of a solution. It is defined as 0 MPa for STP (absolute pressure of 1 atm = 0.1 MPa). For a case of a partial vacuum or tension as in transpiration, the pressure potential is 0.

Group 1 10.0

ΔW%

The remaining root crops from Part A of the experiment were placed in a blender. The tissue was pureed. The blended root crops were filtered with cheesecloth to remove cell wall and debris, and was then covered in beaker. 60mL sap was poured in 250mL Erlenmeyer flask with magnetic stirring bar and the thermometer was inserted. The flask was surrounded with ice-salt bath and was stirred vigorously. After the temperature was 1oC, the temperature was recorded every 10 seconds. The data was recorded and the temperature versus time graph was constructed. The true freezing point and solute potential was then determined. The pressure potential of the cells of the root crops were then calculated.

y = -25.887x + 6.4495 R² = 0.9664

0.0 -10.0 -20.0 0

0.2

0.4

0.6

Sucrose Concentration (m)

Figure No. 1

0.8

Group 2

Group 6

y = -10.714x + 10.678 R² = 0.1031

10.0

ΔW%

ΔW%

20.0 0.0

0.0 -10.0 -20.0

-20.0 0

0.2

0.4

0.6

0

0.8

ΔW%

0.0

ΔW%

y = -12.785x + 6.0472 R² = 0.8886

-5.0 0

0.2

0.4

0.6

0

0.8

0.4

0.6

0.8

Figure No. 7

Group 8

Group 4 y = -8.477x + 4.7047 R² = 0.4158

0.0

y = -10.55x + 4.9304 R² = 0.7815

10.0

ΔW%

ΔW%

0.2

Sucrose Concentration (m)

Figure No. 3

5.0

0.8

y = -16.021x + 4.7864 R² = 0.9451

10.0 5.0 0.0 -5.0 -10.0

Sucrose Concentration (m)

10.0

0.6

Group 7

Group 3 5.0

0.4

Figure No. 6

Figure No. 2

10.0

0.2

Sucrose Concentration (m)

Sucrose Concentration (m)

5.0 0.0 -5.0

-5.0 0

0.2

0.4

0.6

0

0.8

0.4

0.6

0.8

Figure No. 8

Figure No. 4

Group 9

Group 5

y = -24.726x + 14.466 R² = 0.9475

20.0

ΔW%

y = -29.788x + 9.7058 R² = 0.959

20.0 10.0 0.0 -10.0 -20.0

0.2

Sucrose Concentration (m)

Sucrose Concentration (m)

ΔW%

y = -23.375x + 6.3734 R² = 0.9811

10.0 0.0 -10.0

0

0.2

0.4

0.6

Sucrose Concentration (m)

Figure No. 5

0.8

0

0.2

0.4

0.6

Sucrose Concentration (m)

Figure No. 9

0.8

𝑅 is the gas constant = 8.31J/ K.mol and 𝑇 is the temperature in Kelvin (28⁰C +273).

Group 10 y = -38.37x + 17.386 R² = 0.9874

ΔW%

20.0 0.0 -20.0 0

0.2

0.4

0.6

0.8

Root Crops Potato

Sucrose Concentration (m)

Carrot Figure No. 10 As seen in Figures 1-10, it is constant for all groups that ΔW% is inversely proportional to sucrose concentration (m). This indicates that as the concentration of sucrose increases, the change in the weight of the root crop decreases. Some of the plots have low coefficient of determination (R2) (Figures 2&4). Inaccuracies can be caused by improper blotting, errors in the analytical balance; leaving cylinders in the solutions for different amounts of time and excessive drying out of the cylinders. The molality where the percent change in weight (ΔW%) is equal to 0 represents the molality at which there is no net movement of water molecules and no osmotic gradient, creating an osmotic equilibrium. Since there is no net osmotic movement, the turgor pressure, which pushes the plasma membrane against the cell wall of the plant sample, does not exist (Steudle & Zimmermann, 2016). Therefore 𝛹𝑝 = 0.

Table No. 1 Water Potentials of Different Root Crops As seen in table 1, the crops have different molalities where ΔW% = 0. The results range from 0.28676 to 0.5509 m; wherein camote has the highest and the radish having the lowest. The relationship of solute concentration (in molality) to solute or osmotic potential is given by the van 't Hoff equation 𝛹𝑠 = −𝑚𝑖𝑅𝑇 where 𝑚 is the molality (1 molal=1 x103 mol/m3 H2O), 𝑖 is the ionization constant = 1 for sucrose,

Jicama

Camote

Radish

Group

Molality of sucrose where ΔW%= 0 (m)

Water Potential (MPa)

Grp. 1

0.2886

-0.7220

Grp. 6

0.2822

-0.7059

Grp. 2

0.4397

-1.0999

Grp. 7

0.3275

-0.8191

Grp. 3

0.4857

-1.2148

Grp. 8

0.4906

-1.2271

Grp. 4

0.5440

-1.3608

Grp. 9

0.5509

-1.3779

Grp. 5

0.2676

-0.6693

Grp. 10

0.4046

-1.0121

The negativity in osmotic potential is due to the fact that pure water is usually defined as having an osmotic potential (𝛹𝑠 ) of zero and the presence of solutes always make a solution have less water than the same value of pure water. Thus, the more solute molecules present, the more negative the osmotic potential is. Typical values for cell cytoplasm are –0.5 to –1.0 MPa. The resulting solute potential in J/m3 was converted into MPa by dividing 106. It is assumed that 𝛹𝑚 is very small and therefore negligible, thus, 𝛹 = 𝛹𝑠 . As seen in table 1, the water potential varies from -0.6693 to – 1.3779 MPa; wherein camote has the highest and the radish having the lowest. A more negative water potential indicates that the cells of the root crop contains more solute. Among the root crops, the cells of the camote have the most solute and the cells of the radish have the least solute. Solute potential is defined as the pressure that must be applied to a solution to prevent the inward flow of water through a semipermeable membrane. The flow of water stops because the pressure of the solution and the pressure of the water are equal. (Martin and Hine, 2008). Solute potential values are always negative. The reason for the negativity of solute potential is due to the

Cryoscopy is a method used for the determination of the lowered freezing points produced in liquid by dissolved substances in order to determine the molecular weight of solutes and various properties of solutions. With this method, the solute potential of extracted sap in plants can be acquired using the formula:

� = degrees of supercooling (negative in sign)
 0.0125 = amount of water (1/80) that solidifies per degree of supercooling.

0

Temperature (oC)

fact that pure water, which is determined to contain no solutes, has a solute potential equivalent to zero. By adding solutes, the solution would result to have less water than the same value of pure water. Therefore, it can be said that the more solute molecules present, the more negative the Solute potential is.

𝛹𝑠 = 1.10Δf𝑐

In determining freezing point depression on tissue sap, increase of temperature would be a direct result from the release of the heat of fusion with sudden increase as the apparent freezing point (Δf') (Saupe, 2009). According to Reiss (1996), after obtaining the correction for supercooling, the true freezing point (Δf) is obtained using the equation: Δf = Δf' - 0.0125 � Where: Δf = true freezing point
 Δf' = apparent freezing point


200

300

-0.5 -1

Time (s)

Figure 1.1. Depression of the Freezing Point of Potato (Group 1)

1.5

Temperature (oC)

Furthermore, cryoscopy centers upon the phenomenon known as Depression in Freezing Point. The freezing point of a substance is the temperature at which a substance’s liquid and solid form is in equilibrium. This phenomenon dictates that the freezing point of a pure solvent would be higher (less negative) than that of the freezing point of a solution. It is also a colligative property wherein it does not depend on the nature of the solute molecules but rather it depends only on the amount of solute molecules present.

100

-1.5

1 0.5 0 -0.5

0

50

-1

100

150

200

Time (s)

Figure 1.2. Depression of the Freezing Point of Potato (Group 6) 0

Temperature (oC)

Where in ‘1.10’ is the ratio of the osmotic potential of a non-iodized solution (-2.27 MPa) and the freezing point depression of sucrose (2.06ºC) (Saupe, 2009), Δf or 𝑇𝑓 is the true freezing point and c is the correction factor or the ratio between the room temperature in Kelvin (28⁰C +273K) and 273 K.

0

-0.5 0

100

200

300

400

-1 -1.5 -2 -2.5 -3

Time (s)

Figure 1.3. Depression of the Freezing Point of Carrot (Group 2)

2

100

200

300

400

500

-1

0

300

-3

Time (s)

Figure 1.7. Depression of the Freezing Point of Camote (Group 4) 1.5

0

1

0

100

200

300

-1 -2

0.5 0 -0.5

0

200

-1

Time (s)

Figure 1.5. Depression of the Freezing Point of Jicama (Group 3)

400

600

800

Time (s)

Figure 1.8. Depression of the Freezing Point of Camote (Group 9)

2

1.5

1 0

50

100

150

200

250

-2 -3

Temperature (oC)

1

0

0.5 0 -0.5

0

100

200

300

400

-1 -1.5

-4 -5

200

-2

1

-1

100

-5

Time (s)

-3

Temperature (oC)

0

-4

Figure 1.4. Depression of the Freezing Point of Carrot (group 7)

Temperature (oC)

Temperature (oC)

1

Temperature (oC)

Temperature (oC)

2.5 2 1.5 1 0.5 0 -0.5 0 -1 -1.5 -2

-2 Time (s)

Figure 1.6. Depression of the Freezing Point of Jicama (Group 8)

Time (s)

Figure 1.9. Depression of the Freezing Point of Radish (Group 5)

Table 2. Determination of Freezing Point of Different Root Crops

Temperature (oC)

1.5 1 0.5

Root Crops

Group

Apparent Freezing Point (Δf') -1oC

True Freezing Point (Δf)

Solute Potential (Ψs)

Grp. 1

Degree of Supercooling -1.4oC

Potato

-0.9825oC

-1.1915 MPa

Grp. 6

-0.7oC

-0.7oC

-0.6913oC

Grp. 2

-2.4oC

-2oC

-2.375oC

Grp. 7

-1.6oC

-1.4oC

-1.5825oC

Grp. 3

-2.7oC

-1oC

-2.6875oC

Grp. 8

-4.2oC

-2.4oC

-4.17oC

Grp. 4

-4oC

-1.2oC

-3.985oC

Grp. 9

-0.6oC

-0.3oC

-0.5963oC

-0.8384 MPa -2.88 MPa -1.9193 MPa -3.2595 MPa -5.0575 MPa -4.833 MPa -0.7232 MPa

Grp. 5

-1.4oC

-1.3oC

-1.3838oC

Grp. 10

-1.2oC

-0.9oC

-1.1888oC

0 -0.5 0

100

200

300

400

-1 -1.5

Time (s)

Figure 1.10 Depression of the Freezing Point of Radish (Group10) In this experiment, the concept of supercooling is applied. Supercooling is defined as the process where in a liquid reaches is freezing point but does not crystallize. This process occurs because supercool liquids are metastable and do not possess seeds which triggers the crystallization of the liquid sample. In Figures 1.1, 1.3, 1.5, 1.6, 1.7 and 1.10, the lowest point in the graph corresponds to the extent of the supercooling process happening within the sap solution, the abrupt increase from the lowest point to its next point is termed as the “apparent freezing point”. This abrupt increase as shown in the stated figures is because of the release of the heat of fusion. However, in figures, 1.2, 1.4, 1.8 and 1.9 and the theoretical increase cannot be seen, this may be because the extent of the supercooling process of the root crop was not efficiently carried out. It may also be because of human error. The extent of super-cooling of a solution is the lowest temperature value obtained. After rapid increase in temperature, which is a result of the heat of fusion, the value obtained is the apparent freezing point (Δf') (Saupe, 2009). Using the equation, Δf = Δf' - 0.0125 �, the true freezing point for each of the root crops were obtained. The values of true freezing point for each root crop were used to calculate their solute potentials respectively. All the obtained results are presented in Table No. 2.

Carrot

Jicama

Camote

Radish

By obtaining the solute potential, we can calculate the pressure potential for each root crop with the formula: 𝛹𝑝 = 𝛹 − 𝛹𝑠 The resulting values for the pressure potential is shown in Table No. 3. Table No. 3 Pressure Potential of Different Root Crops Root Crops

Group

Potato

Grp. 1 Grp. 6 Grp. 2 Grp. 7 Grp. 3 Grp. 8 Grp. 4 Grp. 9 Grp. 5 Grp. 10

Carrot Jicama Camote Radish

Pressure Potential 0.4695 MPa 0.1325 MPa 1.7801 MPa 1.1002 MPa 2.0447 MPa 3.8304 MPa 3.4722 MPa -0.6547 MPa 1.0089 MPa 0.4294 MPa

Possible sources of error in determining the true freezing point may be because of the state of the ice used. There were periods within the experiment where in other groups used melted ice while others used new ice.

-1.6782 MPa -1.4415 MPa

Conclusion The water potentials of Potato ranges from -0.70 to -0.72 MPa, Carrot ranges from 0.82 to -1.10 MPa, Jicama ranges from -1.21 to 1.23 Mpa, Camote ranges -1.36 to -1.38 MPa, and Radish ranges from -0.67 to -1.01 MPa. Camote has the lowest water potential and highest solute concentrations while Radish has the highest water potential and lowest solute concentrations. Upon obtaining the water potential and solute potential, they were used to calculate the pressure potential for each of the root crops. Pressure potential for camote ranges from -0.6547 MPa to 3.4722 MPa, for potato it ranges from 0.1325 MPa to 0.4695 MPa, for carrot it ranges from 1.1002 MPa to 1.7801 MPa, for Jicama it ranges from 2.0447 MPa to 3.8304 MPa and for Radish it ranges from 0.4294 MPa to 1.0089 MPa. Application 1. Why is the osmotic potential always negative in sign? Answer: The value of osmotic potential is always negative in sign since the presence of solutes, which water binds to through hydrogen bonding consumes some of the potential energy available in water. Thus, it will always make a solution have less water than the same value of pure water.

2. How do dissolved solutes contribute to the water potential of the cell? Answer: Water potential is defined by its equation: ψ = ψ𝑠 +ψ𝑝 + ψ𝑚 , where in ψ𝑠 stands for the concentration of the solutes or the osmotic potential and it always has a negative sign because of the presence of solutes. Solutes can be dissolved by water because water can bind through them via hydrogen bonding, this process consumes some of the potential energy available in water and since it becomes more negative as more dissolved solutes are added this also lowers the water potential of the cell.

Answer: During equilibrium, the sign of the pressure potential of tissues is 0. At incipient plasmolysis, the pressure potential will still be 0 and during transcription, particularly in xylem cells, there will be a negative pressure potential. 4. What causes supercooling? Why is the lowest temperature reached not considered the freezing point of the sap solution? Answer: Supercooling, also known as undercooling, is the process when a liquid reaches below its freezing but does not become solid. This occurs because supercooled liquids that are metastable do not possess seeds that may trigger the crystallization. The reason why the lowest temperature of the sap solution is not considered because sap solutions usually reach 7 to -16 degrees Celsius, once they reach below their supercooling limit, ice forms and the plant dies (Korner, 2012).

5. Do plants with watery sap have higher or lower water potential? Explain your answer using actual computed values. Answer: Based on the table presenting the computed values for water potential for each root crops, it can be inferred that the more watery sap a root crop has the lower its water potential. This is based on the principle that the higher the concentration of solute a root crop has the more negative its water potential shall be. Root Crops Potato

Carrot

Jicama 3. What is the sign of the pressure potential of tissues at equilibrium? (+), (0), or (-)? At incipient plasmolysis? During transpiration?

Camote

Radish

Group

Molality of sucrose where ΔW%= 0 (m)

Water Potential (MPa)

Group 1

0.2886

-0.7220

Group 6

0.2822

-0.7059

Group 2

0.4397

-1.0999

Group 7

0.3275

-0.8191

Group 3

0.4857

-1.2148

Group 8

0.4906

-1.2271

Group 4

0.5440

-1.3608

Group 9

0.5509

-1.3779

Group 5

0.2676

-0.6693

Group 10

0.4046

-1.0121

7. References [1] Bland, W. and Tanner, C. (1985). Measurement of the Water Potential of Stored Potato Tubers. Plant Physiology. 79: 891-895 [2] Cryoscopy (Theory) : Physical Chemistry Virtual Lab : Chemical Sciences : Amrita Vishwa Vidyapeetham Virtual Lab. (n.d.). Retrieved 27 February 2017, from http://vlab.amrita.edu/?sub=2&brch=190&sim= 337&cnt=1 [3] Experimental Explanation of Supercooling: Why water does not freeze in clouds. (n.d.). Retrieved 27 February 2017 from Science Daily: https://www.sciencedaily.com/releases/2010/04/ 100421133114.htm [4] Haynie, Donald T. (2001). Biological Thermodynamics. Cambridge: Cambridge University Press. pp. 130–136. ISBN 0-52179549-4. [5] Hine, R., & Martin, E. (2008). A dictionary of biology (6th ed.). Oxford: Oxford university press. [6] Korner, C. (2012). Alpine Treelines: Functional Ecology of the Global High Elevation Tree Limits. London:Springer Basel. [7] Kramer, Eric; David Myers. (2012). "Osmosis is not driven by water dilution". Trends in Plant Science. 18 (4): 195– 197. doi:10.1016/j.tplants.2012.12.001. [8] Moller, I., Murphy, A., Taiz, L. & Zeiger, E. (2015). Plant Physiology and Development. (6th ed.). USA:Sinauer Associates. [9] Oxley, A., Spell, E. (2012). Molar Mass Determination by Depression of the Freezing Point. [10] Saupe, S. (2009). Measuring Water Potential by the Gravimetric Technique. Plant Physiology. Collegeville, MN: College of St. Benedict Retrived 27 February 2017 from https://employees.csbsju.edu/ssaupe/biol327/Lab /water/water-lab-grav.htm [11] Steudle, E., & Zimmermann, U. (2016). Effect of Turgor Pressure and Cell Size on the

Wall Elasticity of Plant Cells. Plant Physiology, (170). Retrieved March 5, 2016, from http://www.plantphysiol.org/content/59/2/2 85.full.pdf