Calculation of water potential

Calculation of water potential

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For a mesophyte growing in a moist soil, water potential of its root sap is -3 atm, then how to find the water potential of moist soil and leaf respectively?

I don't know if we can calculate the exact values. The only thing I can say is that the potential of the soil will be less negative than -3 while that of the leaf will be more negative.

You know that water travels from higher potential to lower potential. So, it will travel from a region of less negative water potential to a region of more negative water potential.

Investigation: Osmosis and Water Potential

In this lab, you will observe the process of osmosis and diffusion. You will also learn how to calculate water potential. If you are not familiar with these concepts, make sure that you have looked them up in your textbook. If you don't know what these terms mean, this lab is not going to make sense to you.

  • Investigate the processes of osmosis in a model of a membrane system
  • Investigate the effect of solute concentration on water potential as it relates to living plant tissues

Solute Potential

Solute potential (Ψs), also called osmotic potential, is negative in a plant cell and zero in distilled water. Typical values for cell cytoplasm are –0.5 to –1.0 MPa. Solutes reduce water potential (resulting in a negative Ψw) by consuming some of the potential energy available in the water. Solute molecules can dissolve in water because water molecules can bind to them via hydrogen bonds a hydrophobic molecule like oil, which cannot bind to water, cannot go into solution. The energy in the hydrogen bonds between solute molecules and water is no longer available to do work in the system because it is tied up in the bond. In other words, the amount of available potential energy is reduced when solutes are added to an aqueous system. Thus, Ψs decreases with increasing solute concentration. Because Ψs is one of the four components of Ψsystem or Ψtotal, a decrease in Ψs will cause a decrease in Ψtotal. The internal water potential of a plant cell is more negative than pure water because of the cytoplasm’s high solute content (Figure 2). Because of this difference in water potential water will move from the soil into a plant’s root cells via the process of osmosis. This is why solute potential is sometimes called osmotic potential.

Plant cells can metabolically manipulate Ψs (and by extension,Ψtotal) by adding or removing solute molecules. Therefore, plants have control over Ψtotal via their ability to exert metabolic control over Ψs.

Figure 2. A semipermeable membrane between two aqueous systems

In Figure 2, water will move from a region of higher to lower water potential until equilibrium is reached. Solutes (Ψs), pressure (Ψp), and gravity (Ψg) influence total water potential for each side of the tube (Ψtotal right or left ), and therefore, the difference between Ψtotal on each side (ΔΨ). (Ψm, the potential due to interaction of water with solid substrates, is ignored in this example because glass is not especially hydrophilic). Water moves in response to the difference in water potential between two systems (the left and right sides of the tube).

Practice Question

Positive water potential is placed on the left side of the tube by increasing Ψp such that the water level rises on the right side. Could you equalize the water level on each side of the tube by adding solute, and if so, how?

Water Potential: Measurements, Methods and Components

In this article we will discuss about:- 1. Subject-Matter of Water Potential 2. Measurement of Water Potential 3. Methods 4. Components 5. Water Potential in Cells 6. Movement of Water from Cell to Cell.

Subject-Matter of Water Potential:

In recent years the term chemical potential of water is replaced by water potential. This is designated by the Greek letter psi (Ψ). Water potential is measured in bars. The latter is a pressure unit. When the water potential in a plant cell or tissue is low the latter is capable of absorbing water.

On the other hand, if the water potential of the cell tissue is high it indicates their ability to make available water to the desiccating surrounding cells. Clearly water potential is used as a measure to determine whether the tissue is under water stress or water deficit.

It needs mentioning that it is the difference between the water potential in a system under study and that in a reference state which is taken as the water potential value.

The reference state is pure water at the temperature and atmospheric pressure comparable to that of the system being investigated. As will be clear from Fig. 6-2, the water potential in the reference state is arbitrarily taken a value of 0 bar. The same figure also shows range of Ψ in the different tissues. As will be observed herbaceous leaves of mesophytes have water potentials ranging from —2 to —8 bars.

When the water decreases in the soil the water potential tends to become more negative than —8 bars. It may be added that if the water potential falls beyond —15 bars, most plant tissues stop growing.

The response of herbaceous and desert-growing plant leaves vary when the water potential falls below —20 to —30 bars. Similarly seeds and pollen or spores are having very low water potentials and the values may be as low as —60 to —100 bars.

Measurement of Water Potential:

In studies concerning plant water relations, information on water potential in plant cells and tissues is very vital. Several methods are used to measure water potential but none of them is infallible.

Methods of Water Potential:

Some of the methods are given below:

i. Vapour Equilibrium Method:

Here the pressure of water vapour in equilibrium with the water in a tissue sample enclosed in a small chamber is measured.

The water vapour pressure is measured with the help of thermocouple psychrometer. This is an accurate method to measure tissue water potential.

Some of these psychrometers can measure the water potential of attached leaves up to ± 1 bar.

ii. Vapour Immersion Method:

This method is based on the fact that when a plant tissue is placed in an atmosphere in which water vapour is maintained at constant vapour pressure, there will be a net transfer of water between the tissues and the surrounding atmosphere till an equilibrium is reached.

The difference in the water potentials of the tissue and the environment will determine the quantity of water transferred.

iii. Liquid Immersion Method:

Usually two methods are employed and these are the liquid immersion and dye methods. The former is similar to the vapour immersion method. In general, dye method has several advantages.

iv. Pressure Chamber Method:

Using pressure chamber water potential can be measured within minutes. Further compared to other methods, no precise temperature controls are needed.

The apparatus is also relatively less expensive. This method is especially suited for field studies.

Components of Water Potential:

Keeping in view that a typical plant cell is made up of a vacuole, a cell wall and the cytoplasm between the two, usually three major sets of internal factors are visualized which contribute towards water potential (Fig. 6-3).

These are shown below:

From the given equation it may be inferred that water potential in a plant cell is equal to the sum of the matric potential (ΨM) which is due to the binding of water molecules to protoplasmic and cell wall contents, the solute potential (Ψs Ψπ) due to the dissolved solutes in the vacuoles and lastly the pressure potential (Ψp) which is due to the pressure developed within cells and tissues. These potentials like the water potential are expressed in terms of bars.

In the followings brief accounts of the three components of water potential are given:

Matric refers to the matrix. It is the force of adsorption with which some water is held over the surface of collodial particles in the cell wall and cytoplasm.

It is also written in negative values. In the young cells, seeds and cells of xerophytes its value is appreciable. In the cells of mesophytic plants this is nearly —0.1 atm.

In such instances matric potential is often ignored since it does not contribute significantly to the total water potential.

Accordingly sometimes the equation is modified as below:

This refers to the potential developed by the solute particles in a solution. It is equal to the osmotic potential. Solute potential depends upon the number of particles. In fact, solute potential has replaced the old term osmotic pressure.

The difference is that while the former is expressed in bars with a negative, the latter is written as positive bars. Accordingly when the solute potential decreases it attains more negative value. Several methods are used to measure solute potential in an extracted cell sap. One of these is through the usage of thermocouple psychrometer. Solute potential values vary in plant cells from different species.

iii. Pressure Potential:

This is the hydrostatic pressure which develops in a plant cell due to the inward flow of water: (Ψp). It is also referred to the turgor pressure. Environmental conditions greatly influence the volume, water content, water potential and pressure potential of a cell. In a plasmolysed plant cell, the turgor pressure is zero.

Thus water potential equals the solute pressure or negative osmotic pressure. Or the other hand, in the fully turgid state, the water potential of the cell is zero. At this moment, pressure potential or turgor pressure is equal to solute pressure. Currently very few methods are available to measure pressure potential.

Figure 6-3. A summary diagram showing relationship of different potentials in a cell having elastic walls.

Water Potential in Cells:

The concepts developed on the basis of artificial systems using sugar solution can be successfully transferred to a cell (Fig. 6-4).

Cell is enclosed by a semipermeable membrane and osmosis takes place across this membrane. If a cell is immersed in a solution having high Ψπ (e.g. pure water or a dilute solution), water will diffuse in the cell and the latter will become turgid.

The external solution is referred to as hypotonic solution. In a situation where cell is immersed in a solution having Ψπ equal to its cell sap, no net water diffusion would occur and the cell will remain flaccid or lacks turgor. This solution is called isotonic solution. If the concentration of external solution is more than the cell sap, its Ψπ will be lower than that of the cell sap. If a cell is immersed in such a solution (hypertonic), water will diffuse out and the protoplast will pull inside and become plasmolyzed [Fig. 6-4 (C)].

If such a plasmolyzed cell is placed again in a hypotonic solution, it will again become turgid.

Water potential of a cell has two components (e.g., osmotic and pressure potentials) as follows:

When a cell is immersed in water or a solution and comes in equilibrium the water poential of cell (Ψinside) is equal to the water potential outside (Ψ outside):

Ψπ (inside) + Ψp (inside) = Ψ (inside) = Ψ (outside)

Ψ (outside) is also the total of Ψπ (outside) and Ψp (outside). At atmospheric pressure Ψp = 0, therefore Ψ (outside) = Ψπ (outside).

This may also be mentioned as Ψπ (inside) = Ψπ (outside) -Ψp (inside) and this osmotic potential of the cell sap can thus be measured.

Movement of Water from Cell to Cell:

Differences in water potential (∆Ψ) are important for the water movement in and out of the cell. These differences are relevant as compared with the environments. Likewise water moves from cell to cell by diffusing down the water potential gradient between the two cells.

The direction of water movement and the force of movement are linked with water potential in each cell and also on the difference between the water potential of the two cells (Fig. 6-5).

In the instance mentioned below we observe that:

Obviously water will flow from cell B to cell A i.e., towards the lower or more negative water potential.

The value of ∆Ψ is very vital since it is directly proportional to the rate of movement of water from one cell to another.

The rate and amount of water movement is dependent upon the difference in water potential on either side of the membranes.

Osmotic potential of a solution:

The osmotic potential of a solution can be measured with the help of an osmometer (Fig. 6-6). It is a simple apparatus consisting of a cylinder having water tight sliding piston and a differentially permeable membrane at one end.

When it is immersed in pure water (Ψπ = 0) P, the pressure developed in the osmometer is equal and opposite to the osmotic potential at equilibrium.

V = volume of the solution containing a given amount of the solutes

T = temperature as expressed in degree absolute

R = gas constant (solute molecules freely diffuse as if they were a gas, the constants K1 and K2 can be replaced by it).

Water movement as explained on the basis of old approach to osmosis:

For a long time osmosis was explained on the basis of water diffusion from a zone of high concentration to the lower concentration (diffusion pressure deficit: DPD).

However, this is not correct since some of the solutions occupy a volume smaller than the same weight of pure water.

It was also believed that a solution in a cell was as if sucking water into the cell by a force regarded as a negative pressure.

Several terms w ere used to explain these concepts. In recent years several of these terms have been discarded and more acceptable explanations based on thermodynamic concepts have been advanced. Terms used currently and their old equivalent corresponding terms are given in Table 6-1.

Lab 9 Transpiration Example 2 ap

Most of the water a plant absorbs is not used for a plant’s daily functioning. It is instead lost through transpiration, the evaporation of water through the leaf surface and stomata, and through guttation, which is the loss of water from the vascular tissues in the margins of leaves.

There are three levels of transport in plants: uptake and release of water and solutes by individual cells, short distance cell to cell transport at tissue and organ levels, and long distance transport of sap by xylem and phloem at the whole plant level. The transport of water is controlled by water potential. Water will always move from an area of high water potential to an area with low water potential. This water potential is affected by pressure, gravity, and solute concentration.

Water moves into the plant through osmosis and creates a hydrostatic root pressure that forces the water upward for a short distance, however, the main force in moving water is the upward pull due to transpiration. This pull is increased by water’s natural properties such as adhesion and cohesion. Transpiration decreases the water potential in the stele causing water to move in and pull upward into the leaves and other areas of low water potential. Pressure begins to build in the leaves, so to prevent downward movement, guttation occurs. Guttation occurs through leaf openings on the leaf margins called hydrathodes. Loss of water through transpiration can be facilitated by the opening and closing of the stomata depending on environmental conditions.

There are three types of cells in plants: parenchyma, sclerenchyma, and collenchyma. Parenchyma cells are the most abundant and are not specialized. They are found in the mesophyll of leaves, the flesh of fruits, the pith of stems, and the root and stem cortex. Sclerenchyma are elongated cells that make up fibers. They have thick secondary walls and the protoplasts often die as they grow older. They are used for support and are found in vascular tissue. Collenchyma cells are living at maturity and have a thickened secondary wall.

In Lab 9A, all of the plants in this experiment will lose water through transpiration, but those affected by the heat sink and the fan will lose a larger amount of water due to the environmental conditions. This transpiration will pull water from the potometer into the plant. The structure and cell types of a stem cross-section can be observed under a microscope.

Exercise 9A: Transpiration

The materials needed for this exercise were a pan of water, timer, a beaker containing water (heat sink), scissors, 1-mL pipette, a plant cutting, ring stand, clamps, clear plastic tubing, petroleum jelly, a fan, lamp, spray bottle, a scale, calculator, and a plastic bag.

Exercise 9B: Structure of the Stem

The materials needed for this exercise were a nut-and-bolt microtome, single-edge razor blade, plant stems, paraffin, 50% ethanol, distilled water, 50% glycerin, toluidine blue O stain, a microscope slide and cover slip, pencil, paper, and a light microscope.

Exercise 9A: Transpiration

The tip of the pipette was placed in the plastic tubing and they were submerged in a tray of water. Water was drawn into the pipette and tubing until no bubbles were left. The plant stem was cut underwater and inserted into the plastic tubing. Petroleum jelly was immediately placed around the tube edging to form an airtight seal around the stem. The tubing was bent into a “U” shape and two clamps were used on the ring stand to hold the potometer in place. The potometer was allowed to equilibrate for ten minutes.

The plant was exposed to a fan, which was placed one meter away and set on low speed. The time zero reading was recorded and then it was continually recorded every three minutes for 30 minutes. After the experiment, all the leaves were cut off the plant and massed by cutting a one cm2 box and massing it.

Exercise 9B: Structure of the Stem

A nut-and-bolt microtome was obtained and a small cup was formed by unscrewing the bolt. The stem was placed in the microtome and melted paraffin was poured around the stem. The paraffin was allowed to dry and the excess stem was cut off. The bolt was twisted just a little and then cut with the blade. The slice was placed in the 50% ethanol. The slices were left in the ethanol for five minutes. Using the forceps, the slices were moved to a dish of the toluidine blue O stain and left for one minute. The sections were rinsed in distilled water. The section was mounted on the slide with a drop of 50% glycerin. A cover slip was placed over the slide. The cross section was observed under a light microscope and drawn.


Osmosis is the process in which water flows from an area with a low solute concentration, to an adjacent area with a higher solute concentration until equilibrium between the two areas is reached. [6] In addition, water flows from areas of low osmolarity to areas of high osmolarity. [7] All cells are surrounded by a lipid bi-layer cell membrane which permits the flow of water in and out of the cell while also limiting the flow of solutes. In hypertonic solutions, water flows out of the cell which decreases the cell's volume. When in a hypotonic solution, water flows into the membrane and increases the cell's volume. When in an isotonic solution, water flows in and out of the cell at an equal rate. [4]

Turgidity is the point at which the cell's membrane pushes against the cell wall, which is when turgor pressure is high. When the cell membrane has low turgor pressure, it is flaccid. In plants, this is shown as wilted anatomical structures. This is more specifically known as plasmolysis. [8]

The volume and geometry of the cell affects the value of turgor pressure, and how it can have an effect on the cell wall's plasticity. Studies have shown how smaller cells experience a stronger elastic change when compared to larger cells. [3]

Turgor pressure also plays a key role in plant cell growth where the cell wall undergoes irreversible expansion due to the force of turgor pressure as well as structural changes in the cell wall that alter its extensibility. [9]

Turgor pressure within cells is regulated by osmosis and this also causes the cell wall to expand during growth. Along with size, rigidity of the cell is also caused by turgor pressure a lower pressure results in a wilted cell or plant structure (i.e. leaf, stalk). One mechanism in plants that regulate turgor pressure is its semipermeable membrane, which only allows some solutes to travel in and out of the cell, which can also maintain a minimum amount of pressure. Other mechanisms include transpiration, which results in water loss and decreases turgidity in cells. [10] Turgor pressure is also a large factor for nutrient transport throughout the plant. Cells of the same organism can have differing turgor pressures throughout the organism's structure. In higher plants, turgor pressure is responsible for apical growth of things such as root tips [11] and pollen tubes. [12]

Dispersal Edit

Transport proteins that pump solutes into the cell can be regulated by cell turgor pressure. Lower values allow for an increase in the pumping of solutes which in turn increases osmotic pressure. This function is important as a plant response when under drought conditions [13] (seeing as turgor pressure is maintained), and for cells which need to accumulate solutes (i.e. developing fruits). [14]

Flowering and reproductive organs Edit

It has been recorded that the petals of Gentiana kochiana and Kalanchoe blossfeldiana bloom via volatile turgor pressure of cells on the plant's adaxial surface. [12] During processes like anther dehiscence, it has been observed that drying endothecium cells cause an outward bending force which led to the release of pollen. This means that lower turgor pressures are observed in these structures due to the fact that they are dehydrated. Pollen tubes are cells which elongate when pollen lands on the stigma, at the carpal tip. These cells grow rather quickly due to increases turgor pressure. These cells undergo tip growth. The pollen tube of lilies can have a turgor pressure of 0–21 MPa when growing during this process. [15]

Seed dispersal Edit

In fruits such as Impatiens parviflora, Oxalia acetosella and Ecballium elaterium, turgor pressure is the method by which seeds are dispersed. [16] In Ecballium elaterium, or squirting cucumber, turgor pressure builds up in the fruit to the point that aggressively detaches from the stalk, and seeds and water are squirted everywhere as the fruit falls to the ground. Turgor pressure within the fruit ranges from .003 to 1.0 MPa. [17]

Growth Edit

Turgor pressure's actions on extensible cell walls is usually said to be the driving force of growth within the cell. [18] An increase of turgor pressure causes expansion of cells and extension of apical cells, pollen tubes, and in other plant structures such as root tips. Cell expansion and an increase in turgor pressure is due to inward diffusion of water into the cell, and turgor pressure increases due to the increasing volume of vacuolar sap. A growing root cell's turgor pressure can be up to 0.6 MPa, which is over three times that of a car tire. Epidermal cells in a leaf can have pressures ranging from 1.5 to 2.0 MPa. [19] As plants can operate at such high pressures, this can explain why they can grow through asphalt and other hard surfaces. [18]

Turgidity Edit

Turgidity is observed in a cell where the cell membrane is pushed against the cell wall. In some plants, their cell walls loosen at a quicker rate than water can cross the membrane, which results in a cell with lower turgor pressure. [3]

Stomata Edit

Turgor pressure within the stomata regulates when the stomata can open and close, which has a play in transpiration rates of the plant. This is also important because this function regulates water loss within the plant. Lower turgor pressure can mean that the cell has a low water concentration and closing the stomata would help to preserve water. High turgor pressure keeps the stomata open for gas exchanges necessary for photosynthesis. [10]

Mimosa pudica Edit

It has been concluded that loss of turgor pressure within the leaves of Mimosa pudica is responsible for the reaction the plant has when touched. Other factors such as changes in osmotic pressure, protoplasmic contraction and increase in cellular permeability have been observed to affect this response. It has also been recorded that turgor pressure is different in the upper and lower pulvinar cells of the plant, and the movement of potassium and calcium ions throughout the cells cause the increase in turgor pressure. When touched, the pulvinus is activated and exudes contractile proteins, which in turn increases turgor pressure and closes the leaves of the plant. [20]

As earlier stated, turgor pressure can be found in other organisms besides plants and can play a large role in the development, movement, and nature of said organisms.

Fungi Edit

In fungi, turgor pressure has been observed as a large factor in substrate penetration. In species such as Saprolegnia ferax, Magnaporthe grisea and Aspergillus oryzae, immense turgor pressures have been observed in their hyphae. The study showed that they could penetrate substances like plant cells, and synthetic materials such as polyvinyl chloride. [21] In observations of this phenomenon, it is noted that invasive hyphal growth is due to turgor pressure, along with the coenzymes secreted by the fungi to invade said substrates. [22] Hyphal growth is directly related to turgor pressure, and growth slows as turgor pressure decreases. In Magnaporthe grisea, pressures of up to 8 MPa have been observed. [23]

Protists Edit

Some protists do not have cell walls and cannot experience turgor pressure. These few protists are ones that use their contractile vacuole to regulate the quantity of water within the cell. Protist cells avoid lysing in solutions by utilizing a vacuole which pumps water out of the cells to maintain osmotic equilibrium. [24]

Animals Edit

Turgor pressure is not observed in animal cells because they lack a cell wall. In organisms with cell walls, the cell wall prevents the cell from lysing from high-pressure values. [1]

Diatoms Edit

In Diatoms, the Heterokontophyta have polyphyletic turgor-resistant cell walls. Throughout these organisms' life cycle, carefully controlled turgor pressure is responsible for cell expansion and for the release of sperm, but not for things such as seta growth. [25]

Cyanobacteria Edit

Gas-vaculate [ check spelling ] cyanobacterium are the ones generally responsible for water-blooms. They have the ability to float due to the accumulation of gases within their vacuole, and the role of turgor pressure and its effect on the capacity of these vacuoles has been observed in varying scientific papers. [26] [27] It is noted that the higher the turgor pressure, the lower the capacity of the gas-vacuoles in different cyanobacterium. Experiments used to correlate osmosis and turgor pressure in prokaryotes have been used to show how diffusion of solutes into the cell have a play on turgor pressure within the cell. [28]

When measuring turgor pressure in plants, many things have to be taken into account. It is generally stated that fully turgid cells have a turgor pressure value which is equal to that of the cell and that flaccid cells have a value at or near zero. Other cellular mechanisms taken into consideration include the protoplast, solutes within the protoplast (solute potential), transpiration rates of the plant and the tension of cell walls. Measurement is limited depending on the method used, some of which are explored and explained below. Not all methods can be used for all organisms, due to size and other properties. For example, a diatom doesn't have the same properties as a plant, which would place constrictions on what could be used to infer turgor pressure. [29]

Units Edit

Units used to measure turgor pressure are independent from the measures used to infer its values. Common units include bars, MPa, or newtons per square meter. 1 bar is equal to .1 MPa. [30]

Methods Edit

Water potential equation Edit

Turgor pressure can be deduced when total water potential, Ψw, and osmotic potential, Ψs, are known in a water potential equation. [31] These equations are used to measure the total water potential of a plant by using variables such as matric potential, osmotic potential, pressure potential, gravitational effects and turgor pressure. [32] After taking the difference between Ψs and Ψw, the value for turgor pressure is given. When using this method, gravity and matric potential are considered to be negligible, since their values are generally either negative or close to zero. [31]

Pressure-bomb technique Edit

The pressure bomb technique was developed by Scholander et al., reviewed by Tyree and Hammel in their 1972 publication, in order to test water movement through plants. The instrument is used to measure turgor pressure by placing a leaf (with stem attached) into a closed chamber where pressurized gas is added in increments. Measurements are taken when xylem sap appears out of the cut surface and at the point which it doesn't accumulate or retreat back into the cut surface. [33]

Atomic force microscope Edit

Atomic force microscopes use a type of scanning probe microscopy (SPM). Small probes are introduced to the area of interest, and a spring within the probe measures values via displacement. [34] This method can be used to measure turgor pressure of organisms. When using this method, supplemental information such as continuum mechanic equations, single force depth curves and cell geometries can be used to quantify turgor pressures within a given area (usually a cell).

Pressure probe Edit

This machine was originally used to measure individual algal cells, but can now be used on larger-celled specimens. It is usually used on higher plant tissues but wasn't used to measure turgor pressure until Hüsken and Zimmerman improved on the method. [35] Pressure probes measure turgor pressure via displacement. A glass micro-capillary tube is inserted into the cell and whatever the cell exudes into the tube is observed through a microscope. An attached device then measures how much pressure is required to push the emission back into the cell. [33]

Micro-manipulation probe Edit

These are used to accurately quantify measurements of smaller cells. In an experiment by Weber, Smith and colleagues, single tomato cells were compressed between a micro-manipulation probe and glass to allow the pressure probe's micro-capillary to find the cell's turgor pressure. [36]

Negative turgor pressure Edit

It has been observed that the value of Ψw decreases as the cell becomes more dehydrated, [31] but scientists have speculated whether this value will continue to decrease but never fall to zero, or if the value can be less than zero. There have been studies [37] [38] which show that negative cell pressures can exist in xerophytic plants, but a paper by M. T. Tyree explores whether this is possible, or a conclusion based on misinterpreted data. In his paper, he concludes that by miscategorizing "bound" and "free" water in a cell, researchers that claimed to have found negative turgor pressure values were incorrect. By analyzing the isotherms of apoplastic and symplastic water, he shows that negative turgor pressures cannot be present within arid plants due to net water loss of the specimen during droughts. Despite his analysis and interpretation of data, negative turgor pressure values are still used within the scientific community. [39]

Tip growth in higher plants Edit

A hypothesis formed by M. Harold and his colleagues suggests that tip growth in higher plants is amoebic in nature, and isn't caused by turgor pressure as is widely believed, meaning that extension is caused by the actin cytoskeleton in these plant cells. Regulation of cell growth is implied to be caused by cytoplasmic micro-tubules which control the orientation of cellulose fibrils, which are deposited into the adjacent cell wall and results in growth. In plants, the cells are surrounded by cell walls and filamentous proteins which retain and adjust the plant cell's growth and shape. As explained in the paper, lower plants grow through apical growth, which differs since the cell wall only expands on one end of the cell. Further more studying on this topic is put on hold due to the pandemic. [40]

AP Biology

To examine molecules in the process of diffusion through a selectively permeable membrane. This experiment is to not only test diffusion but to also test and explore the permeability of a membrane with dependent variables of solute sizes.

In order to understand this section of the lab experiment, it's important to understand the basic concepts that are dealt in the laboratory. Diffusion is defined as the concept of when particles move from an area of higher concentration to an area of lower concentration which is coined with the term concentration gradient. This transportation or movement requires no energy is called with the term passive transport.
A simple application to this with human experience is when one sprays a bottle of air freshener in a room. Molecules of the air freshener will slowly move out and pervade in the room because there is a higher concentration of air freshener molecules in the bottle than there is in the room before it was sprayed.

A selectively permeable membrane is when a membrane permits these molecules or particles (specifically smaller ones) to pass and move through but will be more restricting in terms of allowing larger particles to pass through. The movement through the membrane of these particles is called dialysis.


Left Beaker: Distilled H2O
Right Beaker: Iodine
Potassium-Iodine (IKI)
Dialysis Tubes filled with
15% glucose, 1% starch
((Before)) Dialysis tubes were placed in each

After 30 minutes, noticeable color
change in the dialysis tube in the
beaker with the IKI
((After)) Dialysis tube taken out from the IKI solution.
Dialysis tube is now a
dark greenish color

Based off of this experiment, there are certain substances that were entering and leaving the dialysis tubes because of the selective permeability of the membrane that only allows certain molecules to diffuse. First of all, glucose has came out of the dialysis tubes and into the IKI and water based on the testapes that were both dipped in the beaker with the solution water and IKI. The color of the testapes had a greenish color which meant there was a very small concentration of glucose present in the IKI and water which meant that glucose was small enough to pass and diffuse through the pores of the membrane. Starch remained in the dialysis tubes while the substance IKI entered the dialysis tube because of the evident transformation of color (from colorless to a greenish-dark blue color). This meant that starch did not diffuse into the beaker with IKI or else it would've turned the same greenish-dark blue color. From this result, this meant that starch was too big of a molecule to diffuse through the pores of the dialysis tube's membrane. Water, which was was smaller than the pore size, has also diffused into the dialysis tube because of the concentration of glucose and starch was higher than the concentration of the water molecules that were in the dialysis tube. As a result, water diffuses inside of the dialysis tube to enter dynamic equilibrium of high concentration to low concentration. IKI was also trying to reach a dynamic equilibrium because there is less IKI solution in the dialysis tubes filled with 15% glucose, 1 & starch so IKI moves into the dialysis tube that has a low concentration of IKI. IKI was also able to move in through the pores because of its small size.
Theoretically, if this experiment was modified and designed to be an experiment that shows that water diffused into the dialysis bag by using quantitative data, one would have to measure the mass of the dialysis tube from the beaker that is filled with distilled water. If one didn't know that whether or not water diffused into the tube, one would put the dialysis tube on the scale before it's submerged in the beaker filled with distilled water and then afterwards take it out and weigh it again. If there is an increase in mass of the dialysis tube, that means that water has diffused and entered into the tube. Ranking the size of the pores of the membrane, water molecules, glucose molecules, IKI molecules, and starch molecules, it would be from smallest to largest water molecules, glucose, IKI, membrane pores, then starch molecules. Water, glucose, and IKI molecules are smaller than the pore membranes because they were able to diffuse through the membrane while the starch molecules are the biggest because they could not diffuse through the membrane. If the experiment began with a glucose and IKI solution inside the bag and with starch and water outside, the results one would expect would be that glucose and IKI would diffuse out of the dialysis tube and into the beaker with water and starch. The final solution color inside of the beaker would be greenish-dark blue while the dialysis's tube would not change because starch cannot diffuse into the dialysis tube.
The results of the experiment support of what I thought the outcome would be where I didn't exactly know which molecules could enter initially, but the outcome showed the fact that pore size is relevant to what comes and goes inside of a selectively permeable membrane.

Conclusion: The dialysis bag was a membrane that was selectively permeable and only allowed glucose, IKI, and water to diffuse in and out of the membrane because of its relatively smaller size than the pores of the membrane.

Purpose: To examine the process of osmosis by looking at the relationship between solute concentration and movement of water in a selectively permeable membrane. It is to explore what would happen to the movement of water if the solute concentration (molarity) increases.

Osmosis is the process in which water molecules diffuse through a selectively permeable membrane from a high water concentration (which means lower solute concentration) to of a low water concentration (higher solute concentration) until the solute concentration reaches equilibrium.

To understand this experiment where osmosis is being investigated, one would have to know tonicity. Tonicity is where a solution affects the cell because it is surrounded by that solution causing that cell to lose or gain water and the process of osmosis will take place.
Isotonic is when referred to two solutions that have the same concentration of solutes. Separated by a selectively permeable membrane, there will be movement of water between the two solutions, but ultimately there will be no net change in the amount of water of each solution.
A solution with LESS solute (more water) is hypotonic to the solution with MORE solute (less water) while a solution with MORE solute (less water) is hypertonic to the solution with LESS solute (more water).

To determine the water potential of potato tuber cells

OSMOSIS Aim To determine the water potential of potato tuber cells. Background knowledge Osmosis is defined as the movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane. Osmosis is considered in terms of water potential and solute potential. Water potential is a measure of the kinetic energy of water molecules. Here, water molecules are constantly moving in a random fashion. Some of them collides with cell membrane, cell wall, creating a pressure on it known as water potential.

The higher heir kinetic energy the more they move and hits the membrane, therefore higher the water potential. Water potential also depends on the number of molecules of water present. Solute restricts the movement of water, so a strong sugar solution with lots of solute particles will lower the kinetic energy and hence water potential. In osmosis water moves down a water potential gradient until the equilibrium is reached between two regions.

The amount of solute molecule that lowers the water potential in a solution is called solute potential.

The highest water potential is found in pure water, where no solutes restricts the movement of water. The water potential of pure water is given a value of O colossally. Adding solutes lowers the water potential to a negative value. The more solute particles added, the lower the water potential becomes which means that the more negative value. For example, Pure water = O Spa (highest water potential) Dilute sugar solution = -kappa Concentrated sugar solution = -kappa (lower water potential) Therefore, the nearer the value to O Spa, the higher the water potential.

Potato tubers are made of plant cells. Plant cells always have a strong cell wall surrounding them. They also contain partially permeable cell membrane. They boot have different way of being permeable to the molecules of water and other substances. Membranes are composed of phosphoric belayed with proteins scattered amongst them. In watery solution the phosphoric molecules naturally forms a belayed, two layers of phosphoric with phosphates on the outside and fatty acids pointing inwards.

Membranes allow selective things to pass through the belayed but prevent other things passing through. Cell membrane allows small molecules such as oxygen, carbon dioxide to pass through easily. Water enters the membrane by osmosis avoiding the hydrophobic centre. Plants have strong cell wall made up of cellulose. When they take up water by osmosis they start to swell, but this cell wall prevents them from bursting as the living part of cell inside the cell wall starts to push against the cell wall.

This inelastic cellulose support makes it possible for the plant cell to be turgid whereas in animal cell if water enters the cell by osmosis, the cell burst as volume increases and pressure builds up against cell membrane, which is unable to cope with the pressure. Plant cell wall is made up of cellulose and has an arrangement of fibers around it. It as a very high tensile strength which makes the cell to withstand the large pressures that develop within it as a result of osmosis. It makes the cell wall very rigid and the arrangement of the fibers around the cell helps it to determine the shape of the cell as it grows.

Cellulose fibers are freely permeable, allowing water and solutes to reach the plasma membrane. Therefore cell wall is permeable to water. Cell surface membrane is partially permeable as it is made of phosphoric belayed. The outside of the layer is attracted to water and the centre of the membrane is lipid allowing specific substances to pass through. This partially permeable membrane allows the free passage of some particles but is not freely permeable to others. Biological membrane are freely permeable to water. Sucrose and starch molecules are large therefore they are not freely permeable to membranes.

They are transported across membranes by special process called facilitated diffusion. Solutions with a lower solute concentration are described as hypotonic solution and solution with higher solute concentration is described as hypersonic solution. If the solution surrounding a cell has the same solute concentration as the cell then water ill pass equally between them. When a solution has the same solute concentration as the cell then it is described as isotonic. Starch is insoluble in water because it is formed when many macroeconomics condense together to form long chains.

As it is long chain polymers and complex carbohydrates, the bonds which hold them together are hard to be broken down by water. Therefore starch is insoluble in water. Sucrose is soluble in water because it is a disaccharide sugar. These sugars are formed when two hexes sugar molecules react together by condensation reaction. When water is added in this sugar, it is Rosen down into hexes sugar (small unit), as bonds holding the disaccharide sugar are spitted up by water. Preliminary Work I did an experiment using sweet potato and sucrose solution. I measured the change in length and the change in mass of the sweet potato.

Both mass change and length change gave me a graph but mass change graph was more reliable than length change. The length of the potato cylinder changes because water enters the cell by osmosis down the concentration gradient and therefore volume of cell is increased. As the concentration of solution is increased, the percentage change in length decreases as solute potential in the solution is increased. To obtain value for solute potential of the sweet potato I have relied on the results of mass change against the concentration as the readings obtained forms a pattern on the graph.

The graph produced for length change is a bit odd. This is because it is hard for a cell to push against the cell wall on all sides and increase the length due to the fact that the cell wall is made up of strong cellulose molecule which has high tensile strength, therefore it is hard for a water to push against this support. The effect of this push caused by osmosis is little than the actual push without cellulose in cell wall. Therefore the results obtained for this will not be accurate enough. Hence, I have decided to rely on the mass change of the potato cell.

It is better to use change in mass because we get reliable readings about the water being entered and increased its mass. Sweet potato has high concentration of solute molecules. Sweet potato cell has sucrose which is soluble in water whereas normal potato has starch which is insoluble in water. Prediction When the potato is left in pure water it will gain mass. I base this prediction on the knowledge that I have about osmosis. Osmosis is term given for the movement of eater when it moves from higher water potential to lower water potential.

In pure water, there is no solute molecule therefore the concentration of water is higher then the potato which have high concentration of solutes than that of water. Therefore there is net movement of water from the region where they are in high concentration (that is in pure water) to region were there is low water potential present (in potato) by osmosis. Hence water enters the potato cell. Therefore there will be increase in the mass of the potato as water enters down the water potential gradient. In pure eater the potato tuber swells because water enters their cells by osmosis.

When the potato is left in a high concentration of sucrose, it will lose its mass. This is because there is large number of solutes molecules present in the solution. Here, the water potential is lower than that of potato cell which has high water potential and low solute concentration. Therefore water will leave the potato cell to reach equilibrium with the sucrose solution. Hence, as water leaves the mass of the potato lowers. In rich sugar solution the potato tubers shrink because water leaves the potato by osmosis. At a certain sucrose concentration, there will be no change in mass.

This is because in the cell of potato and the solution surrounding will have equal concentration of solute and water molecule. Hence there would not be any net movement as the water potentials are equal to each other. Apparatus – 6 clean test tubes -6 potato tubers -MM of sucrose solution -Water as needed to make dilutions -Razor blade -Cork -Test tube stand -Syringe -Electronic balance -Paper towel -Ruler (mm) Procedure ? Make dilutions of sucrose solution according to the volume of water needed to dilute the sucrose solution to make several different concentrations egg 0. M, 0. MM, as shown in below table.

Concentration (M)Volume of 1 M sucrose (ml)Volume of water (ml)Total Volume (ml) 00100150 0. 220801 50 0. 440601 50 0. 66040150 0. 880201 50 1 . 01000150 -Use a syringe to measure out different amounts of sucrose solution and water, (total volume as ml). Then pour it into the test tubes in a percentage ratio which will give various molar concentration. -With a maker pen, write a molar concentration on each test tube as to not confuse yourself whenever handling them. -Take one averaged sized potato and check that it is healthy and hard so that tubers obtained are in good condition to use.

Cut out 6 potato tubers using a borer then measure out the length as mm and cut them with razor blade carefully as it is sharp which might lead to injury, and then weigh them to exactly 0. 01 g accuracy on the electronic balance. – I am using a borer to cut the potato because I want a cylinder which has same width and circumference to ensure that I can use potatoes of same shape and then be able to compare them. -Record the initial mass of the potato cylinder -Place one in each test tube and leave it tort 30 minutes but no more than hours because if left longer there is chance of potato becoming soft to handle.

While waiting, Set out paper towels to dry potato tuber. -After 30 minutes, come back to the experiment and then drain the solution in sink and place each cylinders on the paper towel according to the test tube concentration as to not confuse yourself wondering the concentration of cylinder and then recording wrong mass for the potato tuber. -l then dried the cylinders on the paper towel so that no excess water is left on them, as this might have given increased mass if solution was left on the cylinder. -Each potato cylinder was accurately measured to nearest gram (0. Egg) and masses were recorded.

Graph was plotted of percentage mass change of potato against concentration of sucrose solution. Diagram Results obtained Concentration (M)Mass at start (g)Mass at end (g)Change in mass (g)Change in mass 00. 670. 700. 037. 58 0. 20. 670. 690. 022. 99 0. 40. 670. 680. 01 1 . 50 0. 60. 660. 6600 0. 80. 680. 66-0. 02-2. 94 1 . 00. 670. 61-0. 06-8. 96 00. 660. 710. 054. 48 0. 20. 670. 700. 034. 48 0. 60. 680. 6800 0. 80. 680. 67-0. 01-1 . 47 1 . 00. 630. 60-O. 03-4. 76 Concentration (M)Mass at start (g)Mass at end (g)Change in mass (g)Change in Mass 00. 690. 710. 022. 90 0. 20. 690. 710. 022. 90 0. 40. 700. 10. 01 1 . 43 0. 0. 680. 66-0. 02-2. 94 0. 80. 680. 66-0. 03-4. 41 1 . 00. 700. 63-0. 07-10 Average Concentration(M)average (%) 07. 584. 482. 904. 99 0. 22. 994. 482. 903. 46 0. 41 . 501 . 431 . 431 . 45 0. 600-2. 940. 98 0. 8-2. 94-1 . 47-4. 412. 94 1 +8. 96-4. 76-107. 91 Factors to be controlled *Temperature must be at room temperature. This will make sure that the osmosis is not affected by different temperature. High temperature can increase the rate of osmosis and low temperature can decrease the rate of osmosis. *Using same balance to weigh chip throughout the experiment, as measurement may e slightly different between scales. Same type of potato must be used for tubers as different potatoes have different concentrations. *The size of potato is very important. This is because if the size of the potato varies, so will the amount of mass it can loose or gain . So I am going to try to cut all pieces of potatoes to the same size. Because of the use of Borer to cut potato, I will have cylinders with same width and dimensions but the length will differ. Therefore I have decided to cut it to CM in length so that each cylinder has same length and can be compared with each other to see the effect of osmosis.

The volume of sucrose solution in which potato tubers are immersed must be same. This is because to enable the experiment to be fair and have same effect of volume on the potato. Measurements The mass of potato is variable and will be measured throughout the experiment. The potato chip will be measured before it is put in solution and after. This will show us the difference in mass and whether or not osmosis has taken place and how much osmosis has occurred. The mass be recorded in grams. The sucrose concentration which I am using is 1. MM and to make dilutions, I am going to mix water to this concentration.

I have decided to take 2 sets of results from my classmates, as one set of result may not be accurate because of human error. Variable The only factor I am varying is the concentration of sucrose solution Risk Assessment Although the experiment is fairly safe, we have to be aware of some points. While using razor blades, we have to take extreme care and precision as it is very sharp and could easily cause serious injury. Therefore care should be taken. We have to make sure the space and surface where the experiment is going to be carried out should be dry and tidy as watery surrounding may affect the results .

Any breakage should be reported to teachers straight away as we are handling with glass equipment which are dangerous to handle if broken. Evaluation I HTH k that the experiment went very well, there were no big odd results and they produce a good graph. While I was working on the experiment, all necessary precaution were taken into consideration. I think the procedure used is appropriate enough to make overall conclusion of the investigation. But few problems were faced while doing the experiment. Referring to the graph, I can see that as the concentration of sucrose solution is increased from 0.

M to 0. 6 M the percentage change in the mass of the potato is decreasing. Here I mean that the potato is gaining mass. The trend is that it is a negative correlation and there is a high Jump at every two points e. G between 0. MM . MM. I think I have obtained a bit odd value between the concentration of 0. MM and 0. MM. Between these points there is no high Jump with compared to the trend. This could be due to rate of osmosis at these concentration but to notice it is a short Jump with compared to long ones. Some odd results are seen when I am comparing my results with compared to average value. E odd results are underlined in the average table above. The average value of mass change obtained for MM concentration is 4. 99% but the value obtained from me is far more greater than this I. E. 7. 58%. ‘ think this difference is due to the following reasons: As the averages were found from two sets of result from classmates, I cannot be sure about their reliability. Maybe they have done some inaccuracies in the experiment which would have led us to different values. Or maybe I was inaccurate in my measurements. I think the following are reasons which would have affected the results obtained by me.

In the experiment, the concentration of the solution was prepared by the group of students. Therefore I cannot be so reliable on them for making dilutions. Maybe this might have caused some inaccurate results. To be more reliable, I would do dilutions myself if the experiment is to do again as this is a limitation in the procedure as all students have done the experiment in same way. I cannot be sure about the reliability of the mass of potato when they were dried on the filter paper. Here maybe I have dried them more or maybe I have dried them less. If I have dried them more then there will be slight decrease in mass than the actual ass.

If I have dried them less then there will be molecules of solution on them increasing the mass of potato. This was limiting strategy in the experiment as all student have faced same problem as I have. To improve I will be more careful and patient to dry the potato tuber and if possible I will probably find a nice paper which takes up water on anything precisely without drying the substance. I think I was not precise enough to record the mass of potato tuber. This is because while weighing the potato taken from solutions , I remember that the value on the

Water Potential of a Potato

I have been asked to investigate the water potential of a potato. During my investigation I will not actually be able to measure a value for, as it is a pressure value (kPa).

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Osmosis is a special form of diffusion involving only water molecules. It is defined as “The passage of water from a region of high water potential to a region of low water potential through a partially permeable membrane”

Definition of Water Potential.

The potential for water to move across a selectively permeable membrane, where the osmotic potential of distilled water is 0 (kPa) and any water movement is measured with a negative value.

When a solute is mixed with a solvent the water potential will decrease, and will always be a negative value.

As the presence of a solute lowers the water potential, the more concentrated the solute is, the lower the water potential will be.

Pressure Potential.

This is if pressure where applied to a solution it would resist the movement of water. There is an equation that can represent to relationship of water potential, solute potential and pressure potential:

Like animal cells, plant cells contain a number of solutes. When these cells are put into pure water, water moves into the cell because there is a lower water potential in the cell. But unlike animal cells, plant cells cannot control the amount of fluid surrounding the cell.

The Essay on Tuber Cells Water Potential

. present the lower the water potential such change is referred to as the solute potential. To find the water potential of the cells we need to . their water potential is more negative than that of distilled water, which is zero k Pa. When a cell becomes fully turgid the pressure . stomatal aperture) between them. When the stoma is open the plant takes in CO 2 during the day for photosynthesis. However .

If a plant cell is placed in a solution with a lower water potential than the plant cell, water will move leave the cell, and the cell becomes plasmolysed (flaccid).

If the plant cell is place in a solution with the same water potential, there will be no net movement of water. The cell is then said to be incipient plasmolysis.

The table below shows what happens to the cell when water moves in and out of a cell.

Water potential of solution compared to cell Higher Equal Lower

Net movement of water Enters cell Neither enters or leave Leaves cell

Protoplast Swells No change Shrinks

Condition of cell Turgid Incipient plasmolysis Plasmolysed (flaccid)

A plant cell, unlike an animal will not burst when water moves into a cell, it becomes turgid. This is because the cellulose cell wall prevents it from bursting.

To find the water potential of a potato I will be placing fixed volumes of potato, with the same surface area, into different concentrations of sucrose solutions.

I began by cutting cylinders of potato using a borer which had a diameter of 1cm. I then cut the cylinder of potato to a fixed length of 2cm. I then weighed the mass of the cut potato cylinder using a sensitive balance. I did this twice, so I had 2 cylinders of potato the same volume.

One of these potato cylinders was put into a beaker containing 0.2M sucrose solution and the other in a beaker containing 0.7M sucrose solution. Everyone put their potato cylinders in the same beakers.

They were left in the solution for approx. 48 hours, we then measured the mass of the potatoes again. We were unable to be sure which potato we were measuring the mass of, because they were un-marked. This gave us 2 sets of results for each concentration of sucrose solution. With these results I calculated the mean mass before and after the potato cylinders were put in the solution, the difference between the mean mass and I was able to determine if water entered or left the potato.

0.2M Sucrose Solution 0.7M Sucrose Solution

The Essay on Dconcentrations Of Solutions Determine The Mass Of A Potato

Introduction: The way to get the full results of this lab was through the process of osmosis. Osmosis is the movement of water across a membrane into a more concentrated solution to reach an equilibrium. When regarding cells osmosis has three different terms that are used to describe their concentration. The first of these words is isotonic. Cells in an isotonic solution show that the water has no .

Before After 48hrs Before After 48hrs

Mass of Potato Cylinder 0.87 0.93 0.93 0.9

Mean Mass (g) 0.90 0.94 0.91 0.75

From the results I can see the water moved into the potato after 48hrs in the 0.2M as the mass increased by 0.05g. I also saw that water moved out of the potato in the 0.7M solution after 48hrs as the potatoes mass deceased by 0.16g, it had Plasmolysed.

I have to keep the shape of each piece of potato the same to make the experiment as fair as possible and to ensure that the results I record are accurate and reliable. To do this I will be using a 1cm diameter borer so I am able to cut each piece of potato, and it will have the same diameter each time. I will then cut the length of potato will a sharp scalpel to a fixed length of 2cm, giving me a volume of 1.57cm3. I have decided to use these values because they were used in the prelim experiment and I felt that they worked well.

I am going to make a stock of 1M sucrose solution, that I will dilute using distilled water. From my preliminary result I can see that I do not need to use any sucrose solution below 0.2M or higher than 0.7M concentrations. This is because the prelim results showed that water moved into the potato when it was put in 0.2M solution and water moved out when placed in 0.7M solution. But I am planning to find out at what concentration here is no net movement of water which will be between these to values. The concentrations of solutions I plan to use are 0.3M, 0.45M and 0.6M. I am also planning to place the potato in 30cm3 of sucrose solution.

In my preliminary experiment we left the potato tubes in the solution for 48 hours. I plan to do the same. This is because after a while there will be no net movement of water, as there is no water potential difference and the prelim did show a change in mass.

How I Am Going To Create the Stock Solution?

Amount of Stock required: 60cm3

60cm3 x 0.342g = 20.52g in 60cm3 = 1M solution

The Essay on Potato Chip Concentration Mass Chips

Osmosis Investigation We are trying to find out what the concentration of the solutions is inside the vacuole of a potato cell. We will investigate how the mass of the potato chips change in different sucrose solution concentrations. We will use osmosis to do this. I will be measuring and controlling many different variables, these include -The dependent Variable - Weight of potato chip .

How I Am Going To Create the Different Concentrations:

Concentration (M) 0.3 0.4 0.5 0.6

Volume of 1M stock solution (cm3) 9 12 15 18

Volume of Distilled water (cm3) 21 18 15 12

I plan to put the potato cylinder in a boiling tube will 30cm3 of the different concentrations of solutions. Before putting the potato cylinders in the boiling tube I will weight the mass of the potato tubes, then after the potato tubes have been in the solution for 48 hours I plan to weight the mass again. But before doing this I must blot the potato with a paper towel to remove any water which may alter the mass. Using these values I will calculate the change in mass and will be able to determine in which direction the water moved.

To improve the consistency of my results I plan to put 5 potato cylinder into each boiling tube, so I will be able to calculate a mean mass. I am going to

do this to make my results more consistent, accurate and reliable. To make sure I will be able to distinguish between the different pieces of potatoes I am going to mark the end of each piece of potato with a different colour. This will allow me to me to make sure that I am measuring the mass of the same potato before and after being in the solution.

Using all of my recorded result I will be able to determine at what sucrose concentration would be required for no net water movement and therefore and equal water potential by plotting a graph. Then using that value I will using the NVB Roberts table to determine the water potential.


I predict that the 0.3M will give the smallest change in mass because in the preliminary experiment the 0.2M solution the potato only gained 0.05g, which suggests that water moved into the potato and it has become turgid. The potatoes which were in the 0.7M decreased in mass by 0.16g, so water moved out of the potato, causing plasmolysis of the plant cells.

I will require the following apparatus to complete my experiment:

Borer Allows me to cut cylinders of the same diameter

Sucrose Allows me to create the stock solution

The Essay on Osmosis Of Potatoes In Different Sucrose Solutions

The aim of this experiment is to test whether more water moves out of a potato when it is placed in a sweeter sucrose solution than a potato in a less sweet solution. The hypothesis of the experiment is that we expect more water to move out of the potato placed in the sweet solution than the potato placed in a less sweet solution. Independent variable: concentration of sucrose, concentrations: .

Scalpel Allows me to cut an accurate length of potato

Cutting Tile Stops me from cutting into the work surface

Balance Scales Accurate weighing of the potatoes

Paper Towels Clean up any spillages and remove solution from surface of potato

Boiling Tubes Larger than test tubes so it will allow me put 5 potato cylinders in them

Boiling Tube Rack Allows me to keep the boiling tubes upright.

Stop Clock Accurately allows me to time the 48hrs

Distilled Water Dilute and make stock solution without adding additives that may be in tap water

Measuring Cylinders Finer gradations so more accurate volume can be measured

Permanent Marker To make the potato pieces and not be washed off

I must following safety precautions to minimise the risk of injury:

Clean up any spillages to prevent slipping

When not using the scalpel putting it in a safe place to prevent cutting me or anybody else.

When using the scalpel, use a cutting tile to prevent cutting into the work surface.

Wearing an apron/overalls and goggles.

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Calculation of water potential - Biology

Water Relations of Plant Cells and Tissues

In order to understand plant water relations, we have to understand some basic physical principles of water, and water vapor.

Vapor pressure is the partial pressure of water molecules in the gaseous state. Just as Henry's Law for partial pressures works for other gases, so it applies to water too. Thus, the more water in the air, the greater the vapor pressure that water exerts.

Pure water, if placed in a sealed container, will initially lose some water molecules by evaporation to the airspace above the liquid. Eventually though, the air will become saturated by water, at which point, the rate of evaporation from the surface will just equal the rate of condensation, and the amount of water in the air will remain constant.

When the air is completely saturated above pure water, we say that the air is at 100% relative humidity at that temperature. Relative humidity is defined as:

%RH = The amount of water in the air*100/amount of water the air could hold at that temp.

The kinetic activity of the water in the liquid determines the number of water molecules that escape from the surface and go into the gaseous form. The higher the kinetic activity, the faster the rate at which water molecules evaporate. When substances are dissolved in the water, such as salt or sugar, they cause water molecules to lose kinetic energy, because they are attracted to charged sites on these ions and molecules, effectively immobilizing them, and keeping them from evaporating. This lowers the overall energy state of the water, and fewer molecules evaporate as a result.

This means that the same vapor pressure can not develop over a solution as compared to pure water . Thus, the vapor pressure over a solution will be lower than that over pure water, and hence the humidity too will be lower. This principle can be used to develop calibrations for humidity sensors, as various saturated salt solutions achieve defined humidities. For example, a saturated solution of LiCl yields a RH of about 15%, while NaCl gives a much higher humidity of 75% (See Table E-46 in the 56 th Edition of the CRC Handbook of Physics and Chemistry for other values).

Now, if we were to take two containers of water, and separate them by a semi-permeable membrane (one that allows water to go through its pores, but not solutes like salt or sugar), and add sugar to one side, this would result in a lowering of the kinetic energy of the water-sugar solution. Thus, from a statistical-probability point of view, we would expect the molecules of pure water to encounter the membrane more often than the lower energy water molecules on the solution side, and thus, over time, water will move from the pure water to the solution. Of course some water molecules do go the other way, but the net exchange favors movement into the solution. This is known as osmosis. It is a special case of diffusion.

Diffusion can be defined as the random movement of molecules from an area of high concentration to one of lower concentration. Note the emphasis on random. If pressure is used to move molecules, then the process is known as bulk flow. Fick's Law can be used to express diffusion:

J = - DaC

where J is the flux of material crossing a unit area per unit time (i.e., moles m 2 s -1 )

D is the diffusion coefficient, a proportionality constant that is influenced by the medium in which the molecule
travels, and the molecule itself

a is the area over which the material passes,

l is the length over which the molecule diffuses, and

C is the difference in concentration between the two ends of the path

C is the true driving force for diffusion. If there is no difference, than this term is zero, and no diffusion takes place. The negative sign in the equation simply means that diffusion goes towards the region with lower concentration. C could apply to any gas, including water, and if water, then the difference is one of vapor pressures.

In summary, as Hopkins points out, diffusion is directly proportional to the diffusion coefficient, area, and C, and inversely proportional to the distance moved.

Pressure is simply force per unit area, and in physics is expressed as Newtons / m 2 .

1 atmosphere = 14.7 lbs/in 2 or 760 mm Hg or 1.013 x 10 5 N/m 2

This is equivalent to the pressure of 10.35 meters of water.

In plant physiological work, we don't usually use the above units, but go metric, with either bars or megapascals (MPa).

1 bar = 14.5 lbs/in 2 or 0.9869 atmospheres or 10.22 m of water.

The more solutes that are dissolved in solution, the greater the osmotic pressure that can build up. V'ant Hoff, a chemist in the 18 th century, discovered a mathematical relationship between moles of solute and osomotic pressure, known today as V'ant Hoff's Law:

where pi is the osmotic pressure (let's use bars for now)

n is the number of moles of solute

R is the gas constant (0.08314 liter bar mol -1 K -1 )

T is temperature in Kelvin (K) and

V is the volume of solvent (i.e., water) (liters)

If you re-arrange the above equation, V = nRT/pi , you'll see it is strictly analogous to Boyle's Law, which you learned in chemistry, but now which applies to a liquid situation.

Suppose we dissolved 1 mole of ideal solute into 1000 g of water. What would be the osmotic pressure? Assume a temperature of 1 o C, or 274.16 o K. At this temperature, 1 g of water is essentially 1 cm 3 , and 1000 cm 3 is one liter. Therefore:

pi = (1)(0.08314)(274.16)/1 = 22.79 bars or 2.279 MPa

What if we use salt, NaCl, instead of an ideal solute? Then the osmotic pressure comes out to be 43.2 bars, not 22.79. Why?

Remember, NaCl dissolves into Na + and Cl - , so one mole of solid salt yields two moles of ions!! Substitute n = 2 into the above equation, and you'll get 45.58 bars. This is slightly different from the actual 43.2, and is due to other physical factors which we'll not go into. Suffice it to say, if you want to know the exact osmotic pressure generated by a particular solute, you have to go to tables prepared by researchers who have documented the values exactly.

If you dissolve one mole of sugar into one liter of water, you also don't get 22.79 bars, but rather, something closer to 25.1 bars. Why again does this not match up with the predicted? Turns out sugar, especially glucose, has charges on it that result in a shell of 6 water molecules around the sugar, and this further lowers the kinetic energy of the water molecules, similar to adding more ideal solute. Thus sugar solutions result in higher than predicted osmotic pressures.

Chemical Potential of Water
In this section we shall derive the concept of water potential, a thermodynamic expression of the relative water status of a plant. If you remember back to your introductory chemistry courses, you'll recall the concept of chemical potential. We can use that concept, and adapt it specifically to the situation involving water, to derive a measure of plant water status known as water potential.

Why derive this, when perhaps there are other, simpler ways to express plant water status? Let's go over those, and see why they don't work.

Plant Water Content
One way to express the water content of a plant is to simply weigh the plant, or plant organ, then dry it, and reweigh, and express the water status as percent water weight. But the problem with this is that plants vary in the amount of cell wall material, and other structural materials, such that two different plant species might contain the same water , but have different cell weights , which would lead us to think that their water status differed, when in fact, it might not.

Relative Water Content (RWC)
This is simply a refinement of the above method, but which standardizes water content to the maximum amount of water a plant or plant organ might hold. How is it measured? First, you obtain a plant weight, known as fresh weight. Then you hydrate a plant (put it in standing water, or water it thoroughly) to full turgor (this means it has taken up all the water it can) and you reweigh it. Then, you dry the plant completely, and get the dry weight. Then you express the amount of water in the plant as a percentage of the total amount of water the plant could take up:

RWC (%)= (Fresh Wt. - Dry Wt.)*100/(Turgid Wt. - Dry Wt.)

Since the numerator is equivalent to the fresh water weight and the denominator to total water weight, the equation reduces to:

RWC (%) = Fresh Water*100/Total Water

This works quite well within a particular species, but studies have shown it does not carry over among different species or the same species grown under different conditions. The problem again is that even relative water contents don't mean the same thing to different plants, or even to the same plant under different growth conditions. Thus, researchers in the middle part of this century began a search for an alternative water status parameter which could be applied across all plants, and all conditions. At a meeting of plant physiologists in the early 1960's it was suggested that thermodynamic properties should apply for all plants, and the concept of water potential, which is based in thermodynamics, was born.

Water Potential
The chemical potential of water can be defined as the free energy per mole of water. Chemical potential is, simply put, the potential for a substance to react or move, or in other words, to do work. Work results when a force applied on an object causes it to move from one location to another. Thus, if osmosis results in the movement of water from one location to another, then work has been done. This implies that the chemical potential must be different on either side of the membrane, or else there would have been no potential to do work in the first place. This also suggests that water moves from higher potential to lower potential. As an example, consider that if gravity causes water to run downhill, then gravity has lowered the chemical potential of that water. At the top of the hill, there is abundant potential energy, as the water is supported against the pull of gravity. But, if a gate is opened, and the water runs out, then that potential energy is converted to kinetic energy, and as that kinetic energy moves the water downhill the mean free energy of that water decreases until it reaches a lower stable energy level in the ocean or lake.

Chemical potential depends on the mean free energy of water, and the concentration of water molecules (which chemically is referred to as the mole fraction ).

You already know that solutes act by (1) decreasing the mean free energy of water, and, (2) by decreasing the mole fraction of water (solutes take up room that otherwise would be occupied by water molecules, thus decreasing the density of water in solution). Given this, pure water will have a higher chemical potential than will a solution . We can express this mathematically as the reduction in chemical potential between pure water and that of a solution:

where µw is the chemical potential of a solution
µw o is the chemical potential of pure water
R is the gas constant
T is temperature in K
ln is the natural log symbol
C is the concentration of water

Normally, as your book explains, potential is expressed as a function of chemical activity (a), but for our purposes, we'll substitute concentration. They are closely related: a is approximately equivalent to C (see your book for more details). Taking the ln of a and multiplying it by RT (the right hand term in the equation above) converts chemical activity to energy per mole.

Simply put, the chemical potential of water is that of pure water plus the contribution (usually negative) of whatever solutes are dissolved in it.

We can add another term for the effect of pressure on chemical potential, VP, where V is the partial molal volume of water, or more simply, the volume occupied by 1 mole of water. Since 1 liter of water (1000 g = 1000 cm 3 ) is 55.5 moles, then V is 1000/55.5 = 18 ml/mole, or 18 cm 3 /mole. P is the pressure. Since plants can develop significant pressures, we usually express P as the difference between atmospheric and that in the plant cells. Now we can add the influence of pressure on chemical potential to our equation:

If we substitute the value for osmotic pressure from the V'ant Hoff relationship into our chemical potential relationship, we can express the chemical potential as:

Note here that we have substituted for RTlnC. If we make the assumption that the activity coefficient for pure water is 1, then we see that the osmotic potential of pure water comes out to zero, since the ln of 1 is always zero.

Thus, the difference in chemical potential between pure water and a solution is due to its osmotic and pressure potentials.

To get to water potential, we will need to use relative values, since expressing the absolute chemical potential of water is difficult. Rearranging the above equation, we get:

We use the symbol psi (here written as Y) to symbolize water potential, and therefore:

This is not a trivial usage. It means we can measure the water status (or chemical potential of the water) of a plant by using pressure units, rather than trying to measure the actual energy it would take to move the water (remember, if water moves, energy is being expended).

Normally, we don't use osmotic pressure, but rather, osmotic potential, which is equal to osmotic pressure, but carries a negative sign (to indicate that it lowers the chemical potential of the solution). Pressure potential is positive, since it tends to increase the potential a solution.

Water potential can be broken down into its components as:

where Y w is total water potential,
Y o is the osmotic potential,
Y p is pressure potential,
Y m is matric potential,
g is the potential due to gravitational forces

We have already discussed osmotic and pressure potentials. Matric potential is the lowering of water potential due to the adhesion of water molecules to other molecules on cell membranes. For example, membranes often contain many negatively charged sites, and these will attract and "hold" the polar water molecules. In most situations, the matric potential is small and can be ignored, since only a very small fraction of the total cellular water is affected by matric forces. And for small plants, the contribution to total water potential from gravitational forces is also small (g works out to be -0.1 bar per meter height). Thus, for most purposes, total water potential can be defined from the contributions of just the osmotic and turgor components.

Watch the video: Συγκέντρωση C Διαλύματος 1 (May 2022).