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College Biology II Name: Period: Date: Diffusion and Osmosis AP Biology Advanced Inquiry Lab The following is quoted from the Flinn Scientific publication 11117. Background A cell must be able to transport materials back and forth across its membrane to maintain homeostasis. This movement is regulated because cell membranes are selectively permeable. Selective permeability means that some substances can pass through the membrane while others cannot. Both solutes and solvents may cross the cell membrane. Diffusion is the movement of solute from an area of higher concentration to an area of lower concentration. The mechanism of diffusion is quite simple. Molecules and ions are in constant motion. Since they are always moving (due to the kinetic theory of motion), they will eventually collide with one another. The higher the concentration of molecules, the greater the number of collisions. The collisions cause the molecules to change direction and to spread out until they eventually become uniformly distributed. Even after the molecules are evenly distributed, they are still moving, causing them to collide and redistribute. Molecular motion does not cease when uniform distribution (equilibrium) is reached. Uniform distribution is called a dynamic equilibrium because there is no further net movement of the molecules down a concentration gradient. The term concentration gradient simply describes a difference in concentration across a physical distance. Diffusion is one of the key processes involved in the movement of materials into and out of cells and throughout living systems. Water travels through membranes by a diffusion process known as osmosis. Osmosis is the diffusion of [free] water molecules through a selectively permeable membrane from an area where it is more concentrated to an area where it is less concentrated. The terms hypotonic, hypertonic and isotonic are used to describe the relative concentrations of difference solutions. A hypotonic solution has a 1

Background - Web viewThere will be a net flow of water via osmosis from the hypotonic sided to the hypertonic side to equalize the water concentration or water potential

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Page 1: Background -    Web viewThere will be a net flow of water via osmosis from the hypotonic sided to the hypertonic side to equalize the water concentration or water potential

College Biology II

Name: Period: Date:

Diffusion and Osmosis AP Biology Advanced Inquiry Lab

The following is quoted from the Flinn Scientific publication 11117. Background A cell must be able to transport materials back and forth across its membrane to maintain homeostasis. This movement is regulated because cell membranes are selectively permeable. Selective permeability means that some substances can pass through the membrane while others cannot. Both solutes and solvents may cross the cell membrane. Diffusion is the movement of solute from an area of higher concentration to an area of lower concentration. The mechanism of diffusion is quite simple. Molecules and ions are in constant motion. Since they are always moving (due to the kinetic theory of motion), they will eventually collide with one another. The higher the concentration of molecules, the greater the number of collisions. The collisions cause the molecules to change direction and to spread out until they eventually become uniformly distributed. Even after the molecules are evenly distributed, they are still moving, causing them to collide and redistribute. Molecular motion does not cease when uniform distribution (equilibrium) is reached. Uniform distribution is called a dynamic equilibrium because there is no further net movement of the molecules down a concentration gradient. The term concentration gradient simply describes a difference in concentration across a physical distance. Diffusion is one of the key processes involved in the movement of materials into and out of cells and throughout living systems.

Water travels through membranes by a diffusion process known as osmosis. Osmosis is the diffusion of [free] water molecules through a selectively permeable membrane from an area where it is more concentrated to an area where it is less concentrated. The terms hypotonic, hypertonic and isotonic are used to describe the relative concentrations of difference solutions. A hypotonic solution has a higher concentration of [free] water molecules and a lower solute concentration than a reference solution, while a hypertonic solution contains a lower concentration of [free] water molecules and a higher concentration of solute molecules. Hypertonic and hypotonic solutions therefore represent unequal concentrations of molecules on either side of a permeable membrane. There will be a net flow of water via osmosis from the hypotonic sided to the hypertonic side to equalize the water concentration or water potential. Water will continue to move across the membrane in equal amounts creating dynamic equilibrium. See Table 1 for a comparison of hypotonic, hypertonic and isotonic solutions:

Solution Type Water Concentration Solute ConcentrationHypotonic Higher LowerHypertonic Lower HigherIsotonic Equal Equal

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Page 2: Background -    Web viewThere will be a net flow of water via osmosis from the hypotonic sided to the hypertonic side to equalize the water concentration or water potential

College Biology II

Experimental Overview In this lab we will use dialysis tubing to represent model cells. Dialysis tubing is made of a semipermeable membrane similar to that of the cell. There are two sets of experiments, one will involve using our model cells in solutions of various concentration and solute sizes. Here you will explore the direction of osmosis, either into or out of the cell. The second experiment will evaluate the speed of osmosis as a function of solute concentration

Part 1: Modeling Osmosis and Diffusion

1. Fill out column 3 of table 2 using your knowledge of diffusion to make predictions2. Obtain five plastic cups and label them 1 – 5 using a permanent marker. 3. Obtain five 18-cm pieces of presoaked dialysis tubing. 4. Twist approximately 3 cm of tubing at one end and tie it into a knot. 5. Using a 25ml graduated cylinder, measure 15 ml of the specified solution listed for

Model Cell 1 in Table 2. 6. Open the opposite end of the dialysis tubing by rubbing it together between your

fingertips. 7. Place a funnel in the open end of the dialysis tubing and transfer the solution into the

model cell. 8. Twist and knot the open end of the tubing to seal the bag of dialysis tubing. 9. Gently rinse the dialysis tube bag with distilled water to make sure none of the solution

within the bag has dripped on the outside. Note: this step may be omitted if the bag contains distilled water.

10. Place the dialysis tube bag on a paper towel and gently roll it back and forth to remove any excess liquid.

11. Place a weighing dish on the electronic balance and zero the balance. 12. Measure and record the mass of the dialysis tube. 13. Fill cup number one with approximately 100 ml of the solution listed in the Surrounding

Environment column, row 1 of Table 2. 14. Place the dialysis tube bag in cup 1. 15. After 30 minutes, remove the dialysis tube bag from the cup. Gently roll it back and

forth on a paper towel to remove excess liquid. Measure and record the final mass of the dialysis tube bag.

16. Repeat steps 3 – 14 with the remaining solutions chosen in Table 2.

Analysis of Results Calculate the percent change in mass for each model cell and explain the results in terms of membrane permeability, the nature of the solutes, and the solution concentrations. Identify the solutions as hypertonic, hypotonic or isotonic. Discuss whether any of the experiments provide evidence for the permeability of the membrane with respect to specific solutes.

Percent change = (Final mass – Initial mass) / Initial mass x 100%

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Page 3: Background -    Web viewThere will be a net flow of water via osmosis from the hypotonic sided to the hypertonic side to equalize the water concentration or water potential

College Biology II

Cup No.

Model Cell (inside dialysis tubing)

SurroundingEnvironment(plastic cup)

Prediction – Water will diffuse into or out of the cell

PercentChange inMass

1 Distilled water Distilled Water

2 NaCl, 1 M Distilled Water

3 Sucrose, 1 M Distilled Water

4 Albumin Sucrose, 1 M

5 Distilled Water Glucose, 1 M

Table 2

Results/ Comments about experiment:

Part 2 : The effect of solute concentration on the net amount or rate of diffusion into artificial dialysis cells.

1. Wash out the plastic cups previously used with deionized water. 2. Obtain five additional 18-cm pieces of presoaked dialysis tubing. 3. Twist approximately 3 cm of tubing at one end and tie it into a knot. 4. Using a 25ml graduated cylinder, measure 15 ml of the specified solution lin the first

column for Model Cell 2 in Table 3. 5. Open the opposite end of the dialysis tubing by rubbing it together between your

fingertips. 6. Place a funnel in the open end of the dialysis tubing and transfer the solution into the

model cell. 7. Twist and knot the open end of the tubing to seal the bag of dialysis tubing. 8. Gently rinse the dialysis tube bag with distilled water to make sure none of the solution

within the bag has dripped on the outside. Note: this step may be omitted if the bag contains distilled water.

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Page 4: Background -    Web viewThere will be a net flow of water via osmosis from the hypotonic sided to the hypertonic side to equalize the water concentration or water potential

College Biology II

9. Place the dialysis tube bag on a paper towel and gently roll it back and forth to remove any excess liquid.

10. Place a weighing dish on the electronic balance and zero the balance. 11. Measure and record the mass of the dialysis tube. 12. Repeat steps 3 – 14 with the remaining solutions from table 3. 13. Fill cups 1-5 with approximately 100 ml of deonized water.14. Place the dialysis tube bags in each cup15. Wait 30 minutes!16. After 30 minutes, remove each dialysis tube bag from their cups. Gently roll it back and

forth on a paper towel to remove excess liquid. Measure and record the final mass of the dialysis tube bag. (NOTE this involves a lot of waiting, it will be a lot faster if you do all 5 “cells” at the same time)

Fill in the table with the initial and final measurements. Sucrose Concentration in Dialysis bag

Initial Mass of Cell Final Mass of cell Percent change in mass

0.2 M0.4 M0.6 M0.8 M1.0 MTable 3

Results/ Comments about experiment:

Post Lab report Do your own work. If I notice copied answers both students will get a 5% on this lab report. Answer these questions and incorporate them into your own group lab report. Follow your syllabus guidelines for writing your group lab report .

1) Is the rate of diffusion directly proportional to the solute concentration?

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College Biology II

2) What other variables might influence the rate and direction of osmosis?

3) How would diffusion of a starch solution be different than that of a protein?

4) How could you prove or disprove that a specific solute, such as sucrose, was able to diffuse through a semipermeable membrane?

5) What was the hypothesis tested in part 2 of this lab, when you used 5 concentrations of glucose?

6) Patients with kidney failure need to have their blood filtered through dialysis tubing for many hours each week. Based on what you saw in today’s lab, speculate as to why it takes such a long time.

7) Describe the relationship between the mass change in part 2 and the concentration of solute. Sketch a graph of this relationship.

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