Chapter 3-Movement of Substances Across the Plasma Membrane

Biology | Form 4 | Chapter 3: Movement of Substances across the Plasma Membrane | Experiment

Experiment 3.1

Aim To study the movement of substances across the Visking tubingProblem statement What are the factors affecting the movement of substances across a semi-permeable membrane?Hypothesis The movement of the molecules across a semi-permeable membrane depends on the sizes of the molecules.Variables (a) Manipulated variable: Size of solute molecules

(b) Responding variable: Colour of the solution in Visking tubing(c) Fixed variable: Volume of solutions, time, surrounding temperature

Materials 5% starch suspension, 20% glucose solution, distilled water, Benedict’s solution, iodine solution, Visking tubing and threadApparatus Boiling tube, test tubes, pipette, test tube holder and Bunsen burnerProcedure

1. The Visking tubing is soaked in water until soft. A knot is tied at one end of the Visking tubing.2. The Visking tubing is filled with 10ml of starch suspension by using a pipette.3. The other end of the Visking tubing is tied using a piece of thread. The colour of the solution is recorded.4. The outer surface of the Visking tubing is rinsed with distilled water to remove all traces of glucose of starch in the distilled water.5. A beaker is filled with 300ml of water and 10ml of iodine solution. The colour of the solution is recorded.6. The Visking tubing is then placed into the beaker and left aside for 30 minutes.7. After 30 minutes, the Visking tubing is taken out and transferred into a dry beaker.8. The colour of the solutions in the Visking tubing and the beaker is recorded.9. The presence of glucose in the solutions of the Visking tubing and the beaker is tested using the following steps.

(a) 2ml of each of the solutions is poured into separate test tubes and mixed with 1ml of Benedict’s solotion.(b) The test tubes are then heated in a water bath for 3 minutes. The change in colour is recorded.

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ObservationTest

Content of Visking tubing Contents of beakerInitial colour Final colour Initial colour Final colour

Iondine test Clear Blue black Brown BrownBenedict’s test Blue Blue Blue Orange precipitate

Discussion 1. At the end of the experiment, the solution in the Visking tubing turns blue-black. This shows the both iodine and starch are presence in the Visking tubing.

2. Iodine molecules in the beaker can diffuse throught the semi-permeable membrane of the Visking tubing.3. Starch molecules cannot diffuse through the Visking tubing. As the result, the colour of the solution in the beaker remains yellow.4. The brick red precipitate formed shows that glucose is found in the beaker at the end of the experiment.5. This shows that the Visking tubing has pores big enough for the glucose and iodine molecules to diffuse through but too small for

the starch molecules to diffuse through.Conclusion The Visking tubing is a semi-permeable membrane that only allows substances with small molecules to diffuse through it. The move of

molecules across a semi-permeable membrane depends on their sizes. The hypothesis is acceptable.

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Experiment 3.2

Aim To demonstrate the process of osmosis using a simple osmometer.Problem Statement How does osmosis take place?Hypothesis Osmosis takes place when water moves from a region of low solute concentration to a region of high solute concentration across a semi-

permeable membrane.Variables (a) Manipulated variable: Concentration of solution

(b) Responding variable: The level of sucrose solution in the capillary tube(c) Fixed variable: Surrounding temperature

Materials Visking tubing, 30% sucrose solution, distilled water and threadApparatus Beaker, retord stand and clamp, capillary tube, pipette and marker penProcedure

1. The Visking tubing is soaked in water until soft. A knot is tied at one end of the Visking tubing.2. The Visking tubing is filled with 20 ml of sucrose solution using pipette.3. The other end of Visking tubing is then fitted onto one end of the capillary tube and is tied with a piece of thread.4. The outer surface of the Visking tubing is rinsed with distilled water.5. The Visking tubing is immersed into a beaker of distilled water and the capillary tube is clamped to a retord stand.6. The initial level of the sucrose solution in the capillary tube is marked with a marker pen.7. The level of sucrose solution in the capillary tube is marked again after 30 minutes.8. Steps 1 to 7 are repeated to set up a control experiment. Distilled water is used instead of sucrose solution in Visking tubing.

Observation 1. The final level of sucrose solution in the capillary tube is higher than the initial level.2. The water level in the control experiment remains unchanged.

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Discussion 1. The Visking tubing is a semi-permeable membrane. It has pores that are small enough for water molecules to diffuse through it but not big enough for sucrose molecules to diffuse out of it.

2. As the concentration of the sucrose is higher than that of water in the beaker, water molecules will diffuse into the Visking tubing from the beaker. This causes the level of sucrose solution in the capillary tube to move higher.

3. In the control experiment where the concentration of distilled water in the Visking tubing and the beaker are the same, the water level remains unchanged because the rate of water molecules moving into the Visking tubing is the same as the rate of water moving out of it.

Conclusion The hypothesis is accepted that osmosis is a process by which water moves from a region of low solute concentration to a region of high solute concentration across a selectively-permeable membrane.

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Activity 3.1

Aim To study the wffects of hypotonic, hypertonic and isotonic solutions on plant cells.Materials Onion, 0.5 M sucrose solution, 1.0 M sucrose solution and distilled waterApparatus Light microscope, glass slides, cover slips, scalpel, forceps, dropper, white tile and petri dishProcedure 1. A thin layer of epidermis is peeled off from the inside of an onion scale leaf.

2. The epidermis is placed in a drop of distilled water on a glass silde lebelled A. the glass slide is covered with a cover slip.3. Silde A is observed under the microscope and the epidermal cells are drawn.4. Another piece of epidermis is obtained and placed in a drop of 0.5 M sucrose solution on another galss slide lebelled B. the glass

slide is covered with a cover slip.5. Slide B is observed under the microscope and the epidermal cells are drawn.6. Steps 4 to 5 are repeated using 1.0 M sucrose solution and another glass slide, C to replace the sucrose solution.7. After observing under the microscope, the epidermis in the 1.0 M sucrose solution is transferred into a petri dish filled with distilled

water for a few minutes. The epidermis is again placed in a drop of distilled water on a glass slide labeled D. similar to the glass slides before, glass slide D is covered using a cover slip.

8. Slide D is observed again under the microscope and the epidermal cells are drawn.Observation Slide Observation Shape or condition of the cell

A (distilled water) Turgid, very large vacuole

B (0.5 M sucrose solution) Turgid, vacuole is in the center of the cell

C (1.0 M sucrose solution) Flaccid, small vacuole, cytoplasm shrinks, plasma membrane separates from the cell wall

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D (from 1.0 M sucrose solution into distilled water)

Turgid, very large vacuole

Discussiom 1. The distilled water is less concentrated than or hypotonic to the cell sap of onion cells. When the onion cells are placed in it, water diffuses into the vacuoles of the cells by osmosis. This causes the vacuoles to expand and press against the plasma membrane and the cell wall. The cells are said to be turgid.

2. 0.5 M sucrose solution is isotonic to the cell sap of onion cells. The cells neither gain nor lose water by osmosis.3. 1.0 M sucrose solution is more concentrated or hypertonic to the cell sap of onion cells. Then the onion cells are placed in it, water

diffuses out from the vacuoles of the cells by osmosis. This causes the vacuoles to shrink and the plasma membrane to detach from the cell wall. The cells are filled with water and said to be plasmolysed by the solution and have become flaccid.

4. When the onion cells in the 1.0 M sucrose solution are transferred into distilled water which is hypotonic to water diffuses into the onion cells by osmosis. The cells are filled with water and are said to have undergone deplasmolysis. The cells are turgid.

Conclusion 1. When a plant cell is placed in a hypotonic solution, the cell becomes turgid as water diffuses into it by osmosis.2. When a plant cell is placed in an isotonic solution, the cell neither gain nor lose water by osmosis.3. When a plant cell is placed in a hypertonic solution, the cell becomes flaccid as water diffuses out of it by osmosis. The cell

undergoes plasmolysis.4. When a plant cell placed in a hypertonic solution is transferred back into a hypotonic solution, the cell becomes turgid again as

water diffuses into it by osmosis.

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Activity 3.2

Aim To study the effects of hypotonic, hypertonic and isotonic solutions on animal cells such as red blood cells.Materials Chicken blood, 0.85% sodium chloride solution and distilled waterApparatus Light microscope, glass slides, cover slips, test tubes, pipette and beakerProcedure 1. Four slides are prepared with labels: A, B, C and D.

2. A drop of chicken blood is dropped onto slide A and covered with a cover slip.3. The slide is observed under a microscope and the shape of the red blood cells is drawn.4. A drop of distilled water is dropped onto slide B and covered with a cover slip. A drop of blood is put at one slide of the cover slip.5. When the blood is drawn into the water, the slide is observed under the microscope. The shape of the red blood cells is drawn.6. Steps 4 to 5 are repeated using 0.85% sodium chloride solution on slide C and 5% sodium chloride solution on slide D to replace

distilled water.Observation

Slide A B C DObservation

shape or condition of red blood cell

Biconcave dics shape Fragments of plasma membrane haemolysis

Biconcave dics shape Crinkling of plasma membrane cell(crenation)

Discussion 1. When the red blood cells are placed in the distilled water which is hypotonic, water diffuses into the red blood cells. This causes the cells to swell and burst. As a result, the haemoglobin spills out of the red blood cells. Under the microscope, no red blood cells are observed. There are only broken cell membranes. The red blood cells are said to have undergone haemolysis.

2. The shape of the red blood cells in the 0.85% sodium chloride solution remains the same when observed under the microscope. The red blood cells remain the same as it is placed in an isotonic solution,

3. When the red blood cells are placed in 5% sodium chloride solution which is hypertonic, the water diffuses out from the red blood cells. This causes the cells to shrink. This process is called crenation.

Conclusion 1. In an isotonic solution, red blood cells remain the same.2. In a hypotonic solution, red blood cells burst and undergo haemolysis.3. In a hypertonic solution, red blood cells shrink and crenate.

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Experiment 3.3

Aim To determine the concentration of an external solution which is isotonic to the cell sap of plants.Problem statement Which concentration of the sucrose solution is isotonic to the cell sap of plants?Hypothesis 0.36 M sucrose solution is isotonic to the cell sap of plants.Variables (a) Manipulated variable: Concentration of sucrose solution

(b) Responding variable: Length of potato strips(c) Fixed variable: Temperature, volume of sucrose solution used, time

Materials A potato, distilled water, filter paper, sucrose solutions with concentrations of 0.1 M, 0.2 M, 0.3 M, 0.4 M, 0.5 M and 0.6 MApparatus Seven tested tubes, cork borer, scalpel, forceps, beaker, ruler, glass rod and white tileProcedure 1. Seven cylindrical strips of potato are obtained using a cork borer. Each strip is measured and cut into a length of 5 cm.

2. Seven test tubes were labeled A, B, C, D, E, F and G.3. The test tubes are then filled with the different concentrations of solutions as listed below and placed upright on a test tube rack.

Test tube A: 10 ml of distilled waterTest tube B: 10 ml of 0.1 M sucrose solutionTest tube C: 10 ml of 0.2 M sucrose solutionTest tube D: 10 ml of 0.3 M sucrose solution

Test tube E: 10 ml of 0.4 M sucrose solutionTest tube F: 10 ml of 0.5 M sucrose solutionTest tube G: 10 ml of 0.6 M sucrose solution

4. One strip of potato is then added into each of the test tubes A to G, and left to stand for an hour.5. After an hour, each potato strip is removed from the test tube using the forceps and wiped dry with filter paper. The texture of the

potato strip is observed and the length is measured and recorded in a table.6. From the results obtained, a graph of the change in length of the potato strips aginst the concentration of sucrose solution is plotted.

The concentration at which the change in length is zero is the concentration of the potato cell sap.Observation

Test tube solutionLength of potato strip (cm)

Change in length TextureInitial Final

A distilled water 5.00 5.95 +0.95 TurgidB 0.1 M sucrose solution 5.00 5.60 +0.60 TurgidC 0.2 M sucrose solution 5.00 5.35 +0.35 TurgidD 0.3 M sucrose solution 5.00 5.10 +0.10 TurgidE 0.4 M sucrose solution 5.00 4.95 -0.05 FlaccidF 0.5 M sucrose solution 5.00 4.80 -0.20 FlaccidG 0.6 M sucrose solution 5.00 4.70 -0.30 Flaccid

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Discussion 1. The distilled water an sucrose solutions from 0.1 M to 0.3 M are hypotonic to potato cell sap. Water diffuses into the cells by osmosis and the potato strips becomes turgid and longer.

2. The 0.4 M to 0.6 M sucrose solutions are hypertonic to the potato cell sap. This causes water to leave the potato cell sap by osmosis. The potato strips become mere and more falccid and shorter as the concentration of the sucrose solution increases.

3. The concentration of sucrose solution that is isotonic to the potato cell sap can be obtained by reading from the graph. The concentration at which there is no change in the length in the potato strip is the concentration of the potato cell sap. This is because in an isotonic solution, there is no net movement of water into the cells and out of the cells.

Conclusion The concentration of the potato cell sap is equivalent to 0.36 M sucrose solution. The hypothesis is accepted.

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Chapter 3-Movement of Substances Across the Plasma Membrane