aim of water potential experiment

Practical: Investigating Water Potential ( OCR A Level Biology )

Revision note.

Biology & Environmental Systems and Societies

Practical: Investigating Water Potential

Practical 1: investigating water potential using potato cylinders.

• It is possible to investigate the effects of immersing plant tissue in solutions of different water potentials and then use the results to estimate the water potential of the plant tissue itself
• The most common osmosis practical of this kind involves cutting cylinders of potato and placing them into solutions with a range of different water potentials (usually sucrose solutions of increasing concentration – at least 5 different concentrations are usually required)
• The required number of potato cylinders are cut (one for each of the solutions you are testing – or more than one per solution if you require repeats)
• They are all cut to the same length and, once blotted dry to remove any excess moisture, their initial mass is measured and recorded before placing into the solutions
• They are left in the solutions for a set amount of time (eg. 30 minutes), usually in a water bath (set at around 30 o )
• They are then removed and dried to remove excess liquid
• The final length and mass of each potato cylinder is then measured and recorded

You will need to use apparatus appropriately to measure out the volumes of your solutions and record your measurements

• The percentage change in mass for each potato cylinder is calculated

To find the percentage change in mass, the change in mass must be divided by the initial mass and then multiplied by 100

• The gain of water makes the potato cells turgid, as the water exerts turgor pressure (or hydrostatic pressure) on the cell walls – the potatoes will feel hard
• The potato cylinder in the strongest sucrose concentration will have decreased in mass the most as there is the greatest concentration gradient in this tube between the potato cells (higher water potential) and the sucrose solution (lower water potential)
• More water molecules will move out of the potato cells by osmosis , making them flaccid and decreasing the mass of the potato cylinder – the potato cylinders will feel floppy
• If looked at underneath the microscope, cells from this potato cylinder might be plasmolysed , meaning the cell membrane has pulled away from the cell wall
• If there is a potato cylinder that has neither increased nor decreased in mass, it means there was no overall net movement of water into or out of the potato cells
• The solution that this particular potato cylinder was in had the same water potential as the solution found in the cytoplasm of the potato cells, so there was no concentration gradient and therefore no net movement of water into or out of the potato cells
• The concentration of sucrose inside the potato cylinders can be found if a graph is drawn showing how the percentage change in mass changes with the concentration of sucrose solution
• The point at which the line of best fit crosses the x-axis is the concentration of sucrose inside the potato cylinders

A positive percentage change in mass indicates that the potato has gained water by osmosis (net movement of water from the solution into the potato) meaning the solution had a higher water potential than the potato. A negative percentage change suggests the opposite.

Practical 2: Investigating water potential using onion cells

• If a plant cell is placed in a solution with a  lower water potential  than the cell (such as a concentrated sucrose solution), water will leave the cell through its partially permeable cell surface membrane by osmosis
• As water leaves the  vacuole of the plant cell, the volume of the cell decreases
• The protoplast (living part of the cell inside the cell wall) gradually shrinks and no longer exerts pressure on the cell wall
• As the protoplast continues to shrink, it begins to pull away from the cell wall
• This process is known as  plasmolysis  – the plant cell is  plasmolysed
• Plants with coloured sap (such as red onion bulbs, rhubarb petioles and red cabbage) make observations easier
• The epidermal strips are placed in a  range of molarities of sucrose solution  or  sodium chloride solutions , of gradually decreasing water potential
• Plasmolysis may take several minutes to occur

Light micrograph of normal red onion cells alongside those that have plasmolysed (artistic impression). The cells on the left are epidermal cells that have been immersed in distilled water, whilst the cells on the right are epidermal cells that have been immersed in 1.0 mol dm⁻³ sucrose solution.

Questions involving experiments investigating water potential and osmosis are common and you should be able to use your knowledge of osmosis to explain the results obtained. Don’t worry if it is an experiment you haven’t done – simply figure out where the higher concentration of water molecules is – this is the solution with the higher water potential – and explain which way the molecules move due to the differences in water potential.

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Author: Alistair

Alistair graduated from Oxford University with a degree in Biological Sciences. He has taught GCSE/IGCSE Biology, as well as Biology and Environmental Systems & Societies for the International Baccalaureate Diploma Programme. While teaching in Oxford, Alistair completed his MA Education as Head of Department for Environmental Systems & Societies. Alistair has continued to pursue his interests in ecology and environmental science, recently gaining an MSc in Wildlife Biology & Conservation with Edinburgh Napier University.

• Biology Article
• Study Of Osmosis By Potato Osmometer

Understanding Osmosis Using Potato Osmometer

To study by demonstrating the osmosis process by potato osmometer.

What is Osmosis?

Osmosis is the phenomena in which solvent molecules pass through a semi-permeable membrane from an area of higher concentration to an area of lower concentration. The process continues until the quantity of fluid is balanced or equalized in both regions, the region of higher concentration and the region of lower concentration of the semipermeable membrane. In other words, osmosis is the diffusion or movement of water from a region of higher water potential to a region of lower water potential.

In osmosis, what are solvent and solute?

The fluid that permeates through the semipermeable membrane is called the solvent, whereas the solute is the dissolved particles in the fluid.

What is the solution?

The mixture of solute and solvent form the solution.

List the different types of solutions.

The following are the types of solutions:

• Hypertonic solution – It is a solution with a high solute level. If living cells are placed in a hypertonic solution, because of lower concentration water moves out of the cell causing it to shrink and becomes plasmolyzed.
• Hypotonic solution – It is a solution with low concentration levels of solute. If living cells are placed in this solution, water passes into the cells because of higher water concentration in comparison to the cell causing the cells to swell and turn turgid.
• Isotonic solution – A solution is said to be isotonic if both solutions have an equal concentration of solute. If living cells are placed in an isotonic solution, no change is shown as there is the equal concentration on both the regions hence the cell retains its original shape.

Material Required

• A fresh large-sized potato tuber
• 20% sucrose solution
• Scalpel/blade
• A Bell pin needle that is labelled with a waterproof ink

• Slice the potato tuber into two equal halves with the help of a scalpel or a blade. The outer skin is to be peeled off. Since the tuber shape is irregular, slice the halves into squares
• From the mid-region of the tuber, scoop from the soft parenchyma, so as to form a tiny cavity of a square or a circular shape. At the base, the cavity prepared should have a minimum thickness.
• Fill up half the cavity with the freshly prepared 20% sugar solution. Into the cavity, fix a pin in a way that the mark is in the same line with the layer of the sucrose solution.
• Set up the osmometer in a Petri dish/beaker that is filled with water in a way such that 75% of the potato osmometer is immersed in water
• The set up should remain uninterrupted for close to 1 hour.
• Notice the sugar solution in the osmometer towards the end of the experiment
• Carry out the experiment with the help of water in the cavity and the sucrose solution in the petri dish/beaker.

Observation

After a period of time, within the osmoscope, the sugar solution rises and is seen coloured.

• An increase in the level of sucrose solution is observed in the osmometer. It is because of the entrance of water due to endosmosis from the beaker.
• Also, a water potential gradient is built between the sucrose solution in the external water and the osmometer.
• Though both the liquids are divided by living cells of the potato tuber, they allow the entrance of water into the sugar solution.
• This demonstrates the entrance of water into the sugar solution through the tissues of potato serving as a selectively permeable membrane.

Viva Questions

Q.1. What is plasmolysis?

A.1. It is a process, wherein the protoplasm of the plant cell turns round as a result of contraction when placed in a hypertonic solution due to exosmosis resulting in the decline in the tension of the cell wall.

Q.2. What is the significance of osmosis?

A.2. Osmosis maintains cell turgidity. It causes the transportation of nutrients and discharge of metabolic waste products. It preserves the interior environment of a living entity to maintain an equilibrium between the intracellular fluid levels and water.

Q.3. What is diffusion?

A.3. The movement of molecules from a region of higher concentration to a region of lower concentration. Osmosis is a type of diffusion.

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Water Potential: Measurements, Methods and Components

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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 M ethod:

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 M ethod:

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:

Ψ or Ψ w = Ψ M + Ψ s + Ψ p ; Ψ s = Ψ π

From the given equation it may be inferred that water potential in a plant cell is equal to the sum of the matric potential (Ψ M ) which is due to the binding of water molecules to protoplasmic and cell wall contents, the solute potential (Ψ s ; Ψ π ) due to the dissolved solutes in the vacuoles and lastly the pressure potential (Ψ p ) 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:

i. Matric Potential:

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:

Ψ or Ψ w = Ψ s + Ψ p

ii. Solute Potential:

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 P otential:

This is the hydrostatic pressure which develops in a plant cell due to the inward flow of water: (Ψ p ). 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 Ψ π (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 Ψ π 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 Ψ π 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:

Ψ = Ψ π + Ψ p

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):

Ψ π (inside) + Ψ p (inside) = Ψ (inside) = Ψ (outside)

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

Thus at equilibrium

Ψ π (inside) + Ψ p (inside) =Ψ π (outside)

This may also be mentioned as Ψ π (inside) = Ψ π (outside) -Ψ p (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:

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 K 1 and K 2 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.

Related Articles:

• Experiment to Demonstrate Osmotic Pressure in Plant Tissues
• Mechanism of Absorption of Water | Plant Physiology

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Table of Contents

Investigating Osmosis in Plant Tissue: A Practical Investigation – IGCSE Prescribed Practical

osmosis, plant tissue, sucrose solution, petri dish, concentration, mass, percentage change, water movement, scientific principles

Introduction:

Osmosis is a biological process where water molecules move across a partially permeable membrane from an area of high concentration to an area of low concentration. In this practical investigation, we aim to investigate the process of osmosis in plant tissue. By studying this process, we can gain a deeper understanding of how water moves in and out of cells and how this affects the overall health and growth of plants.

Equipment needed:

• Boiling Tube
• Sucrose solution
• Distilled water
• Paper towel
• Cut the potato into small cubes of equal size using a scalpel and ruler. 1-2cm 3 works well.
• Place the potato cubes in a boiling tube and cover them with a sucrose solution of known concentration.
• Leave the petri dish for an hour and record the initial mass of the potato cubes.
• After an hour, remove the potato cubes from the petri dish and place them on a piece of paper towel to remove any excess liquid.
• Record the final mass of the potato cubes.
• Calculate the percentage change in mass using the formula: Percentage change in mass = ((final mass – initial mass) / initial mass) x 100%

Expected findings:

If the sucrose solution is more concentrated than the potato cells, the water will move out of the cells, and the potato will lose mass. If the sucrose solution is less concentrated than the potato cells, the water will move into the cells, and the potato will gain mass.

Conclusion:

Based on our findings, we can conclude that the concentration of the solution surrounding the plant cells affects the process of osmosis. If the solution is more concentrated than the cells, water will move out of the cells, causing them to lose mass. If the solution is less concentrated than the cells, water will move into the cells, causing them to gain mass.

• What is osmosis?
• How does the concentration of the solution affect the movement of water molecules in osmosis?
• Why is osmosis important for plant growth and health?
• What is the purpose of placing the potato cubes on a piece of paper towel after removing them from the petri dish?
• What would happen to the potato cubes if the sucrose solution was the same concentration as the potato cells?
• Osmosis is the movement of water molecules across a partially permeable membrane from an area of high concentration to an area of low concentration.
• The concentration of the solution affects the direction of water movement in osmosis. If the solution is more concentrated than the cells, water will move out of the cells, causing them to lose mass. If the solution is less concentrated than the cells, water will move into the cells, causing them to gain mass.
• Osmosis is important for plant growth and health because it allows water and nutrients to move in and out of plant cells, which is essential for their survival.
• Placing the potato cubes on a piece of paper towel after removing them from the petri dish is important to remove any excess liquid and prevent it from affecting the mass measurements.
• If the sucrose solution was the same concentration as the potato cells, there would be no net movement of water in or out of the cells, and the mass of the potato cubes would remain unchanged.

By conducting this practical investigation, we can gain a better understanding of the process of osmosis in plant tissue and how it affects the overall health and growth of plants. It is a simple and effective way to study the movement of water in and out of cells and can be used to help explain a wide range of biological processes.

In conclusion, investigating osmosis in plant tissue is an important and informative practical investigation for anyone interested in biology and plant science. By following the steps outlined in this guide, you can conduct your own investigation and gain a deeper understanding of this crucial biological process.

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Test And Quizzes for Biology, Pre-AP, Or AP Biology For Teachers And Students

Lab 9 Transpiration Example 2 ap

Introduction

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.

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.

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.

Table 9.1: Individual Potometer Readings

Class Potometer Readings

Mass of leaves = 1.1 g Leaf Surface Area = 0.0044 m 2

Table 9.2: Individual Water Loss in mL/m2

Table 9.3: Class Average Cumulative Water Loss in mL/m2

Analysis of Results

Calculate the average rate of water loss per minute for each of the treatments: Room: 1.67 mL/m2 Fan: 0.76 mL/m2 Light: 0.93 mL/m2 Mist: 0.83 mL/m2

Explain why each of the conditions cause an increase or decrease in transpiration compared with the control.

How did each condition affect the gradient of the water potential from stem to leaf in the experimental plant?

The light and the fan decreased the water potential in the leaves and water moved up the stem by transpiration pull. The room temperature had little or no effect on the water potential. The mist increased the water potential of the air causing less transpiration to occur from the leaves.

What is the advantage to a plant of closed stomata where water is in short supply? What are the disadvantages?

The closing of the stomata would prevent transpiration of water and minimize this loss if water was in short supply. It is a conservational adaptation. However, closing stomata prevents the exchange of gases in plants and limits their carbon supplies.

Describe several adaptations that enable plants to reduce water loss from their leaves. Include both structural and psychological adaptations.

Plants that are adapted to drier climates are called xerophytes. Some of these plants have adapted small, thick leaves with a reduced surface area. They may also have a thickened cuticle to protect themselves from the environment. The stomata may be sunken into pits. Some xerophytes shed their leaves during the driest seasons and others can store water such as cacti. CAM plants uptake CO2 at night and change it into crassulacean acid that can be broken down during the day for sugars. These plants can close their stomata during the day.

Why did you need to calculate leaf surface area in tabulating your results?

The surface area has to be calculated because this greatly affects the amount of water lost through transpiration. Smaller leaves may lose less water than the larger ones, but by calculated water loss by surface area creates comparable data that is constant and consistent.

Error Analysis

This lab had many opportunities for error. The potometer set up was a complicated procedure. If any air bubbles were present in the plastic tubing, it could cause drastic error to occur. Any miscalculations or inaccurate weighing could also account for error.

Discussion and Conclusion

Transpiration in plants is controlled by water potential. This change in water potential in leaves causes a gradient by which water can be moved upward. When the water potential of the air was increased by the mist and plastic bag, less water evaporated from the leaves, decreasing the water potential gradient between the root and stem. This decreased the transpiration pull. The fan and floodlight simulated environmental conditions such as wind, heat, and intense light. These conditions increase the amount of water transpired by plants. This in turn increased the water potential gradient causing more water to be pulled through the stem. The control plant should have had normal rates of transpiration.

The stem must have specialized cells for support and transport. The epidermis is the outermost layer of the stem. The xylem is a transport tube for water, and the phloem transports food and minerals through the plant. Parenchyma are non-specialized cells and are located in the interior. The tougher sclerenchyma and collenchyma make up the structural outer support of the epidermis and the transport tubes of phloem and xylem.

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The aim of the experiment is to investigate the movement of water particles into and out of potato cells by osmosis and to compare the water potential of old and new potatoes.

An investigation to compare the water Potential of old and new potatoes

From this experiment we should be able to draw conclusions by looking at the change in mass to whether potatoes marketed, as ‘new’ potatoes really are newly grown or just smaller sized old potatoes.

Background Information and Terminology

New Potatoes apart from obviously being smaller in size than old ones have a more distinctive sweeter taste. This is because they are still growing so all their energy is stored as glucose, whereas old potatoes have finished growing and their energy is converted into storage polysaccharides such as starch because it is insoluble.

The water potential ( ψ ) of pure water is zero, when substances are dissolved in water the solutes make the water potential lower so the value becomes more negative. Any substance dissolved in water is called a solute; A solvent is a liquid that is able to dissolve another substance (i.e. A solute) This forms a solution.

The water potential of a solution is the term given to the tendency for water molecules to enter and leave the solution by osmosis. Osmosis is the flow of water molecules by diffusion through a partially permeable membrane, from areas of high water potential (low solute concentration) to regions of low water potential (high solute concentration)

All plant cell membranes are partially permeable, This means that they allow some substances to pass through but not others, whether water enters the cell by osmosis will depend on the balance between external and internal solute and water potentials.

If the solutions on each side of the partially permeable membrane are of equal potential then there will be no movement of water molecules across the membrane. This is called an equilibrium state and the solution is said to be isotonic. A solution that contains more solute particles than the cells is referred to, as being bypertonic while the less concentrated is hypotonic.

Pure water has the highest water potential of zero, If potato discs were placed in pure water, the water molecules would move into the potato cells by the process of osmosis. This would be observed as an increase in size and mass of the potato discs.

With this in mind I expect that as the molarity of the glucose solution increases the potato mass will increase by small increments until it reaches its isotonic point, then the mass will begin to decrease because the solution has a higher concentration of glucose solutes than inside the cell.

This is a preview of the whole essay

I also think that because research shows that new potatoes contain more glucose there will be a greater increase in mass than the old potatoes until it reaches the isotonic point, I predict this because new potatoes will have a higher water potential so more water will move into it.

• 12 lids and specimen tubes
• Ceramic tile
• Measuring Cylinder
• The following molar quantities of glucose solution will be obtained, 0.2, 0.4, 0.7, 0.9, 1.0 and distilled water.
• Using a measuring cylinder 10ml of each molar solution will be measured twice and put in specimen tubes, one for the ‘new’ potato and one for the ‘old’ potato.
• Using a cork borer cut cylinders from a new potato and from an old potato.
• Cut these cylinders to 6 lengths of 10mm, then cut each length into 5 discs of approximately the same size.
• Individually weigh and record each batch of 5 discs then emerse them into the designated glucose solution.
• Leave for 48 hours.
• Once this duration is completed remove the potato from the solution
• Dry off the excess liquid from the discs using paper towel and reweigh recording the mass.
• Independent Variable

The independent variable is the factor which is manipulated, In this experiment it is the concentration of the glucose solution. I chose this for the independent variable because it will be relatively easy to measure different concentrations accurately and successfully.

• Dependant Variable

The dependant variable is what is affected by the independent variable, it is also measurable. The mass of the potato will be the dependant variable because it is affected by the change in concentration (IV) it will be measured using a set of scales.

To ensure my experiment is fair, All variables excluding the independent variable must be kept constant to ensure that the results obtained are as accurate and reliable as possible.

These variables include:

• Temperature - Altering temperature affects the kinetic energy of the molecules, thus affecting the diffusion rate, My experiment will be conducted at room temperature of approximately 22 ° C.
• The length and diameter of the potato, namely 10mm
• The volume of glucose solution will be kept consistent at 10ml.
• The same old and new potato will be used for samples
• All potatoes will be left in the solution for the duration of 48 hours, they must have the same exposure for osmosis to take place.

Risk Assessment

• Care must be taken when using the scalpel
• Care should be taken when using any chemical in the laboratory
• Care should be used with glassware
• Hands should be washed before and after lab-work
• Use the ceramic tile when cutting the potato.

This table shows the results of the experiment involving the new potato.

This table shows the results of the experiment involving the old potato

Using the graphs plotted overleaf, the water potential of both the old and new potato can be calculated. By reading of the isotonic points of both varieties of potato, (this is when there is no change in mass) This concentration can be read off on the calibration graph to tell us the water potential.

The isotonic point of old potatoes  is 0.95 (mol dm )

The isotonic point of new potatoes  is 0.26 (mol dm )

By reading these figures off the calibration graph, the water potentials can be calculated:

The water potential ( ψ ) of old potatoes  is 0.225 (mpa)

The water potential ( ψ ) of new potatoes  is 0.7 (mpa)

Results Analysis

The graph shows that there is a similar pattern for the change in mass for both the old and new potatoes. This being that in distilled water the greatest increases in mass is observed. As the glucose solution becomes more concentrated the increase in original mass rises by smaller increments until no change in mass is observed, after this point the original mass decreases quite rapidly until at around 0.7mol dm  the decrease in mass becomes less drastic.

The general trend is that the greatest increase in mass occurs with distilled water for both old and new potatoes. As the concentration increases, the original mass increases by smaller increments until its reaches its isotonic point. This is an equilibrium state where the water potential of the solution is equal to that of the cell and there is no movement of particles – therefore no change in mass. There is no movement of the water particles, because water moves to regions of higher negativity, if the water potential of the glucose solution is isotonic therefore equal to that of the potato no movement will occur.

As the concentration continues to increase past the isotonic point, the original mass then starts to decrease. This is because there are more solutes present in the glucose solution - The solution is hypertonic, it has more solutes dissolved than the potato and therefore has a more negative water potential. Thus the water particles move from within the potato cell into the solution, observed as a decrease in potato mass.

The greatest increase in mass occurs when the potato is put in the distilled water. The reason behind this is because pure water has a potential of zero because there are no solutes (dissolved molecules) present.

Because water moves into regions of higher negativity molecules will obviously move into the potato cell because water is at zero which is the maximum value so none can move into the water.

This is the general trend for both varieties of potatoes, However what are the individual differences between old and new potatoes?

In distilled water, it is the new potatoes that increase in mass more than the old. The new potatoes increase by 7% more than the old potato; Previous research showed that new potatoes are sweeter because they have glucose energy stores, whereas old potatoes have their energy trapped in storage polysaccharides such as starch. This is simply because it has finished growing and starch is a more efficient storage component because of its secondary structure and insolubility. So therefore because the new potato has a higher value of dissolved solutes it will have a highly negative water potential, therefore more water will move into it so a greater increase in potato mass will be recorded.

Using the calibration graph the water potential of both the old and new potato cells has been calculated to allow for comparison.

The ψ  of old potatoes in 0.225 mpa

The ψ  of new potatoes is 0.7 mpa

The above figures show that new potatoes have significantly higher water potential than old potatoes.

From the conclusions that I have drawn from my results I can conclude that my hypotheses can be accepted that new potatoes have a higher water potential than old because of their composition consisting of higher concentration of glucose energy stores rather than starch.

It also supports the claim that yes potatoes sold as ‘new’ potatoes are actually newly grown potatoes and not smaller sized previously grown potatoes.

Overall the experiment was carried out efficiently and successfully, no real problems occurred. The results obtained were as accurate as could be expected because there are some limitations of accuracy with this experiment. For example the experiment has limited accuracy due to human error and also due to the equipment, measuring the glucose using a measuring cylinder or cutting the potato to size using a ruler has a limited accuracy of ±  0.1mm.

Also it is impossible to tell how much each individual potato piece was dried with paper towel, if some weren’t dried as thoroughly as other this would increase their final weight and could lead to inaccurate misleading results.

However no obvious anomalous results were obtained that could not be fitted into the pattern in some way.

Improvements

• To make the experiment as accurate as possible it needs to be repeated more times, (3 for example) for each concentration. This makes any anomalous results obvious and also for an average to be calculated.
• A greater range of molarities could be investigated, ideally over smaller increments. This would allow for a more precise graph curve to be drawn,
• Ideally all samples should be taken from the same part of the potato to avoid any differences in cell texture and composition.

Bibliography

• Advanced Biology - Principles and Applications C J Clegg & D G Mackean
• Advanced Biology – Simpkins & Williams

Ruth Higgins

Document Details

• Word Count 2002
• Page Count 6
• Subject Science

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Investigation of the Effect of SCA Surfactant on Enhanced Oil Recovery Using Glass Micromodel

• Research Article-Petroleum Engineering
• Published: 05 August 2024

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• Soroush Shojaei 1 &
• Masoud Nasiri   ORCID: orcid.org/0000-0002-5182-6668 1  

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This research investigates the use of a glass micromodel to observe the effects of flooding with a microemulsion on oil recovery and wettability changes. Initially, the internal surface of the micromodel was converted to an  oil wet state to better simulate reservoir conditions. Flooding with distilled water resulted in 75% oil recovery in the  water wet state and 37% in the oil wet state. Surfactant flooding at the optimal concentration of 1 CMC yielded a 57% oil recovery. To further explore the potential of chemical flooding, the study focused on comparing the efficiency of microemulsion flooding to conventional EOR techniques. A Winsor  I microemulsion was prepared using crude oil, sodium cocoyl alaninate, isobutanol, and distilled water. Experiments conducted without salt were successful, recovering 75% of the saturated crude oil. Adding various salts shifted the microemulsion to a Winsor III structure, which is most suitable for EOR. The key factors investigated were the water to oil ratio, cosurfactant to surfactant ratio, and critical micelle concentration. The results demonstrate the superior efficiency of microemulsion flooding, with the best recovery reaching 91%. The findings of this study can inform the design of enhanced oil recovery plans, contributing to more sustainable energy production.

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The authors sincerely thank Dr. Sara Abdi for her cooperation in conducting the experiments and her scientific guidance.

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Shojaei, S., Nasiri, M. Investigation of the Effect of SCA Surfactant on Enhanced Oil Recovery Using Glass Micromodel. Arab J Sci Eng (2024). https://doi.org/10.1007/s13369-024-09327-5

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Teweldebrihan, M.D.; Dinka, M.O. Methyl Red Adsorption from Aqueous Solution Using Rumex Abyssinicus-Derived Biochar: Studies of Kinetics and Isotherm. Water 2024 , 16 , 2237. https://doi.org/10.3390/w16162237

Teweldebrihan MD, Dinka MO. Methyl Red Adsorption from Aqueous Solution Using Rumex Abyssinicus-Derived Biochar: Studies of Kinetics and Isotherm. Water . 2024; 16(16):2237. https://doi.org/10.3390/w16162237

Teweldebrihan, Meseret Dawit, and Megersa Olumana Dinka. 2024. "Methyl Red Adsorption from Aqueous Solution Using Rumex Abyssinicus-Derived Biochar: Studies of Kinetics and Isotherm" Water 16, no. 16: 2237. https://doi.org/10.3390/w16162237

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