At the completion of this exercise you should:
(l) Be able to define the terms diffusion and osmosis.
(2) Be able to list and discuss four mechanisms that cells use to move molecules across their plasma membranes.
(3) Be able to explain what Brownian motion tells us about atoms and molecules.
(4) Be able to explain the relationship between molecular weight and the rate of a molecule’s diffusion.
(5) Be able to list the characteristics of molecules that can, and those that cannot, move passively across a cell’s plasma membrane.
(6) Be able to describe how the solute concentration, inside of a cell, affects the rate (speed) of osmosis.
(7) Be able to define the following terms: concentration gradient, selectively permeable membrane, hypertonic, hypotonic, isotonic, and homeostasis.
Virtually all life forms are composed of cells. The cell is called the fundamental unit of life because within it occur most of the biochemical life processes. One of the phenomena of life is that the chemical composition of a cell remains fairly constant, in spite of the fact that the cell continually uses substances from its external environment and at the same time discharges other substances into its environment: This state of chemical constancy in living systems is called homeostasis. This homeostasis, in addition to the fact that a cell’s surroundings are always of relatively different chemical composition from its inside, leads us to hypothesize that there must be some very selective means of chemical exchange across a cell’s plasma membrane. Today, we will investigate the processes of movement of some substances into and out of cells.
Question 1. Most “cells” do not appear to have an obvious “mouth” or other visible structures in their cell (“plasma”) membranes. Suggest one other way in which materials might be able to pass through the cell’s membrane:
Replace this text with your answer. |
Question 2. Cell biologists tell us that there are 4 basic mechanisms that cells use to get molecules across their membranes. You need to learn these 4 strategies. Go to your textbook, or other reference source, and define the following 4 mechanisms:
1.) | Osmosis | |
2.) | Facilitated diffusion | |
3.) | Endocytosis and exocytosis | |
4.) | Active transport |
Robert Brown made an interesting observation in 1827 that led to the principle that “all atoms and molecules” are in constant motion. Dr. Brown was a botanist and army surgeon who was looking at particles inside pollen grains when he noticed the “rapid oscillatory motion of microscopic particles.” He later observed the same movement when looking at substances, like India Ink. India Ink is made of water and billions of suspended clumps of carbon atoms. Under high magnification, Brown observed that the clumps of carbon atoms were vibrating wildly in all directions. He hypothesized that moving water molecules, which cannot be seen, must be colliding with the clumps of carbon, forcing them to move. Further study has shown that Dr. Brown was correct and, today, we call this kind of observation “Brownian motion”.
A physical scientist would tell you that particles (atoms and molecules) are moving because “they have heat energy (Kinetic energy)”. If you ask what kinetic energy is, you will be told that it is “random molecular motion” which, of course, is a circular argument. The point here is that we really do not know the ultimate reason why all atoms and molecules on earth are moving, only that they are and that the more heat energy (“kinetic energy”) atoms or molecules have, the faster they move.
You are about to make this observation under the compound microscope yourself. The proper way to carry the compound microscope will be demonstrated. Always use two hands. Make sure that the cord is not dangling to prevent a tripping hazard. One hand should be holding the base of the microscope, while the other should hold the arm of the microscope.
Before using the microscope, check for the condition the microscope was left in from the prior class.
Procedure:
Place one edge of the coverslip near to the drop. The stain and water with which you mixed the cells will flow along the junction of the edge of the coverslip and the slide. Carefully lower the coverslip over the specimen keeping the edge of the coverslip in contact with the slide. In this way, the water will flow slowly and uniformly about the specimen and force out air bubbles from beneath the coverslip. (A few air bubbles are not a serious problem for your first slide.)
Looking through the ocular with your right eye only (squint or cover your left eye), bring the specimen into focus by turning the coarse focus adjustment knob slowly until the specimen is generally in focus. Then turning the fine adjustment knob will bring the specimen into sharper focus.
Question 3. Describe the Brownian motion in your own words:
Replace this text with your answer. |
Question 4. Describe, in your own words, how this observation indicates that all visible and invisible molecules are in motion?
Replace this text with your answer. |
Clean up.
Diffusion is the movement of particles from an area of high concentration to an area of low concentration. This results from the continuous random motion that is characteristic of all molecules in liquid or the gas states. A few observations about diffusion will help us to understand how molecules can move from one location to another, perhaps even across cell membranes.
The contents of a cell (the cytoplasm) may be described as a colloid rather than a liquid or solid. Large protein molecules are present in a cell’s cytoplasm that allow it to be in a transitional state of matter called a colloid (somewhere between a liquid and a solid). A special kind of colloid, agar gel, is available in the laboratory and will be used to demonstrate how molecules diffuse from one place to another once they are inside a cell.
Agar is a carbohydrate extracted from algae in powder form. A gel is prepared by mixing the powder with water, then heating followed by cooling–similar to the preparation of a gelatin (animal protein) dessert. The result is a gelatin-like substance composed of intertwined molecules with water trapped among them. Two compounds (molecules), potassium permanganate and methylene blue, have been selected to illustrate diffusion through a colloid. Unlike most components of living cells, these compounds are brightly colored, allowing us to watch their diffusion.
Procedure:
1. Working with your team, obtain a disposable Petri dish containing agar gel.
2. Using a No. 5 cork borer (a “punch”), make two holes in the agar approximately 5 centimeters apart (See diagram below). Remove the plugs of agar with a toothpick and place in the trash.
3. Bring your agar dish, the dropper bottles of potassium permanganate solution and the dropper bottle of methylene blue back to your lab bench.
4. Place two drops of 1% potassium permanganate solution in one of the holes. In the other hole, place two drops of 1% methylene blue solution. Start a timer for 60 minutes. Return the bottles back to the side lab benches for others to use.
5. After approximately one hour, measure the diameter (in millimeters) of the circle that the solutions diffused.
Potassium permanganate: | mm | Methylene blue: | mm |
6. Examine your Petri dish. Shade the agar in the diagram to demonstrate the movement of the solutions in the diagram below. Shade to show where the concentration of the potassium permanganate is higher and where it is lower.
Top view of Petri dish: Side view of Petri dish:
of the agar cut out. Represented by a circle with two smaller circles inside
Right:
Side view of Petri dish with two sections of the agar cut out.” src=”https://docs.google.com/drawings/u/0/d/sw001XxjnioFkNL8IWlK6NA/image?w=428&h=163&rev=1&ac=1&parent=1SUAcijEhcOpgkeRbRil_eOcx-8Rgk2_BZhOOaDXfR_M”>
Question 5. Your Petri dish illustrates the most basic characteristic of diffusion. Complete the following to understand this basic principle:
The net diffusion of potassium permanganate appears to be from the area of its ___ concentration to the area of its _________ concentration.
(“lower to higher” – or – “higher to lower”?)
Question 6. In addition to concentration, the molecular weight of a diffusing molecule is also an important factor that can determine the rate (speed) of diffusion. We happen to know that potassium permanganate’s molecular weight is 150, while that of a methylene blue molecule is 374.
Use your own observations (above) to explain how molecular weight appears to affect the rate of diffusion here:
Replace this text with your answer. |
Question 7. Based on the results, what is the relationship between molecular weight and diffusion rate?
Highlight one: Inverse Direct
State the basis of your conclusion:
Replace this text with your answer. |
Question 8. How would you expect the dyes to be distributed in the agar gel after several days? Would you expect to still see evidence of a concentration gradient?
Replace this text with your answer. |
Question 9. Calculate the molecular weights for the following molecules so you can predict which will diffuse more rapidly.
Note: Molecular weights are calculated by adding up the atomic weights (masses) for all the atoms that make up a given molecule.)Look at the Periodic Table on the wall in the laboratory. It will give you the atomic weights of all the atoms in the two molecules below.
Calculate the weight of these two molecules. Show your work here:
Water, H2O | Sucrose, C12H22O11 | |||||
Number | At. mass | Total | Number | At. mass | Total | |
H: | ||||||
O: | ||||||
C: | ||||||
Total: | Total: |
Question 10. Which substance, water or sucrose, is more likely to be able to diffuse through a semipermeable membrane, and on what basis did you make your decision?
Replace this text with your answer. |
Clean up.
1. Throw away your Petri dish, agar included, into the trash can.
2. Wipe down all countertops and student benches with yellow cleaning solution.
The membranes that surround all cells allow some molecules to diffuse across while inhibiting others. In other words, the membranes of cells are said to be “selectively permeable.” The membranes of all cells are made out of 2 layers of phospholipid molecules with various kinds of channels passing through them. (See Figure 1). Phospholipids are each made of the phosphate-containing “head” which is hydrophilic, and two long fatty acid chains which are hydrophobic.
Question 11. Phospholipids are each made of the phosphate-containing “head” which is hydrophilic (easily combine with water), and two long fatty acid chains which are hydrophobic (do not combine with water, fatty acids are oils). Color the circular phosphate heads blue; color the fatty acid chains of the phospholipids red. Then label them hydrophobic and hydrophilic based on your reading above.
Figure 1. Selectively Permeable Cell Membrane
Question 12. Observe Figure 1.What kinds of molecules are able to pass through the phospholipid bilayer membrane easily? What kinds of molecules are not able to pass through easily? What are the characteristics of each (charge, polar vs. nonpolar), and size?
Move easily | Do not move easily |
Because water is a small and flat molecule (despite it being polar), it can either pass through the phospholipid bilayer membrane. It can more quickly pass into and out of cells through tunnel-like proteins called aquaporins that traverse the cell membrane.
Question 13. Label the aquaporin protein in Figure 1. What does the aquaporin allow to pass through the membrane?
Replace this text with your answer. |
Often, scientists find it helpful to construct a simplified model of an object or phenomenon in order to understand it more clearly. This is exactly what you will do now. You will make several different model cells using a plastic membrane material (dialysis tubing) that mimics the osmosis characteristics of a living cell’s membrane.
The speed or rate at which a molecule diffuses from one area to another depends on the concentration gradient between the two areas. For example, if the concentration of perfume molecules is higher in one room compared to an adjacent room connected by an open door, we would say the concentration gradient between the two rooms is very steep and the rate of diffusion would be very rapid. Conversely, if the concentration of perfume molecules was equal in the two rooms, then the rate of diffusion would be zero and the net movement of the perfume molecules would stop. (Movement would still occur, as Brownian Motion still occurs, but the diffusion would be at equal rates between the rooms.)
Osmosis follows the same laws as diffusion but always refers to water, the principle solvent in cells. A solvent is a fluid that dissolves substances, while the term solute is used to describe substances dissolved in a solvent to make a solution. This means that water will move down its concentration gradient, too, from an area of high concentration of water to an area of low concentration of water.
Because water is the universal solvent in all cells its diffusion into or out of cells is critical in living organisms. Too much water entering or leaving a cell will cause cell death and often the death of the entire organism. You will now set up several different model cells and measure the direction and rate (speed) of osmosis.
Question 14. Define the terms Solvent and Solute:
Replace this text with your answer. |
Question 15. Below is a diagram of three model cells filled with three different solutions. Make a prediction. Review Figure 2 and complete Table 1 to its right. For column 4, predict whether the “model cell” will increase or decrease in weight after one hour. (Hint: Will the net diffusion of water be into the “model cell” or into the beaker?)
Baggy B (representing Cell B) contains 10 mL of 25% sucrose.
Baggy C (representing Cell C) contains 10 mL of 50% sucrose.
The beaker contains DI water (0% solutes).” src=”https://docs.google.com/drawings/u/0/d/sncrbRTGdNJPLGQF5jiTung/image?w=283&h=375&rev=2&ac=1&parent=1SUAcijEhcOpgkeRbRil_eOcx-8Rgk2_BZhOOaDXfR_M”>
Table 1: Prediction for Model Cells
Cell Letter | % solutes in cell | % water in cell | Increase or decrease after 1 hour? |
A | |||
B | |||
C |
Figure 2. Diagram showing the experimental set up for the 3 model cells
Procedure:
5. Open the other end of the tube by rolling with your fingers. Don’t let the tubing get dry, or it may crack.
6. Write down the colors of the solutions here:
1% | 25% | 50% |
7. Fill the 3 plastic tubes with the contents shown below in Figure 2.
8. Twist the other end of the plastic tubing, then fold down and tie a string around that end. The baggie you have created should not be entirely full of water, but instead have a bit of space remaining at the top. Tie the folded ends securely. The goal is to create a water-tight bag.
9. Check for leaks. First, press any extra liquid out of the ends created by filling the bags. Then, by gently but firmly rolling the bag on the paper towel, checking for leaks. Then trim the excess thread. Keep the model cells on the clean paper towel between measurements of cells. Have your instructor check.
10. Check to see if your Triple Beam balance is calibrated properly This means the line on the end of the center beam lines up at the zero line when all of the weights (poises) are slid to the left at zero. If the lines do not meet at the zero line, ask your instructor to assist you.
14. Place cells A, B, and C simultaneously in a beaker filled with 100% distilled water (solvent) (See Figure 2 again).
Note the time here: |
15. Remove all 3 model cells from their beaker every 15 minutes for the next hour. Place the model cells in your paper towel bed. Blot dry and weigh each model cell, being careful to dry the string and the area where you have folded the tubing. Weigh the cells again to the nearest 0.1 gram. Handle the cells very carefully to avoid causing leaks. After you have weighed and recorded the weight, return the “cells” back to the beaker for the next 15-minute period
16. At 15-minute intervals, for 1 hour, record the “total weight” of each cell in Table 2.
17. Calculate the net mass change for each model cell in the last row of Table 2.
Net mass change = Final mass (at time 60) – initial mass (at time 0)
Table 2. Change in Mass of Three Model Cells by Osmosis over Time
Mass (g) | |||
Time (minutes) | Cell A | Cell B | Cell C |
0 | |||
15 | |||
30 | |||
45 | |||
60 | |||
Net mass change |
Question 16. How do your results compare with your prediction in Table 1?
Replace this text with your answer. |
Graph the Osmosis data
1. Read all of the following directions and use google sheets, excel or some other program to graph your data as a line graph. Label the graph with the vertical (“Y”) axis as “Mass (grams)” and the horizontal (“X”) axis as “Time (minutes).”
2. Plot the data from Table 2 for “Mass” at each of the 15-minute intervals.
Replace this text with your graph |
Clean up
Hypotonic, Hypertonic, and Isotonic Environments.
The three terms above are used to describe solute concentration environments a cell may find itself in. In two of these environments water will enter or leave the cell and the cell will change shape and perhaps die. You must learn what these terms mean so you are able to answer questions that use them.
These three terms always describe the solute concentration on one side of cell’s membrane relative to the solute concentration on the other side of the cell membrane.
The term hypotonic means “less solute concentration here” relative to the other side of a cell’s membrane. If you are told the outside environment a cell finds itself in is hypotonic, then you automatically know that the inside environment is the exact opposite or hypertonic.
As you recall, the cell membrane will not allow large or ionic substances to move across it, but water may pass. Your job will always be to determine whether the cell will swell up with water or shrink due to water loss. To do this you must decide which side of the membrane has a higher water concentration. Once this is done you can predict which way water will move (diffuse) since molecules (such as water) always diffuse from an area of higher concentration to an area of their lower concentration.
Look at the osmosis data you collected for Model cell A. We had you put a 1% solute (sucrose) concentration inside the cell and surrounded it with deionized water (a solution that had 0% solute concentration). The proper terminology would be that the inside of the model cell was hypertonic (greater solute concentration) compared to the outside environment that would be described as hypotonic (meaning a lower solute concentration).
To predict which way water will move, you must translate the solute concentrations to water (solvent) concentrations for each side of the cell. Here, the inside of the cell is hypertonic to the outside, hence, the water concentration is lower inside the cell when compared to the outside. So now we can predict that water will move into Model cell A and the cell will get larger. Check your Osmosis data in Table 1 and you should see that, indeed, Model cell A weighed more after 1 hour because it gained water. (If not, you had a leak somewhere!)
Question 17. Do the words hypertonic, hypotonic, and isotonic refer to the solute or solvent concentration?
Replace this text with your answer. |
Question 18. Which solution has the highest concentration of water (the “solvent”): a 1% Sucrose solution or a 50% Sucrose solution?
Replace this text with your answer. |
Question 19. Look at your osmosis data in Table 1:
a. Did the water move into or out of all three model cells? If not, is this what you expected?
Replace this text with your answer. |
b. Would you describe the environment outside the three model cells as hypertonic or hypotonic?
Replace this text with your answer. |
Question 20. A concentration gradient for water must exist between the inside and the outside of a cell’s membrane for osmosis to occur. Observe the graph. Which of the three model cells represents the one with the steepest concentration gradient of water?
Replace this text with your answer. |
Question 21. The steepest concentration gradient of water should result in the highest rate of diffusion (osmosis). Examine your Table 3 osmosis data for the 15 to 30-minute interval. Did the greatest changes in weight occur in the model cells with the greatest concentration gradients of water?
Replace this text with your answer. |
In any liquid that contains various dissolved solutes, all of the different solutes added together constitute the total osmolar concentration (osmolarity) of the solution. It is of critical importance that two solutions which have the same osmolar (solute) concentration also have the same water (solvent) concentration. This is due to the fact that, even though solute molecules may be of different weights, most appear to take up about the same amount of space. The result is that two solutions with different solutes and identical osmolarities also have identical water molecule concentrations (water potentials).
We know that the liquid inside a cell contains many different dissolved substances; they make up the cell’s total osmolar concentration. If a cell is placed in a solution with a higher osmolarity (lower water potential) than inside the cell, we say that the cell is in a hypertonic solution. When a cell is placed in a hypotonic solution, a lower osmolarity (higher water potential) exists outside the cell than inside. A solution that has identical osmolar concentration (water potential) to the cell’s fluid is called isotonic. In this part of the exercise, we will observe the effects of various saline solutions on living animal and plant cells.
Question 22: A cell can be placed in various solutions. Based on the information in the above reading and the concentration gradient, fill in the following table:
A cell placed in a hypertonic solution has … | (more, less, or the same amount of) solutes than the surrounding environment | (more, less, or the same amount of) water than the surrounding environment. | The water (inside of or outside of) the cell will move (into, out of, or equally into and out of) the cell. |
A cell placed in a hypotonic solution has … | (more, less, or the same amount of) solutes than the surrounding environment | (more, less, or the same amount of) water than the surrounding environment. | The water (inside of or outside of) the cell will move (into, out of, or equally into and out of) the cell. |
A cell placed in an isotonic solution has … | (more, less, or the same amount of) solutes than the surrounding environment | (more, less, or the same amount of) water than the surrounding environment. | The water (inside of or outside of) the cell will move (into, out of, or equally into and out of) the cell. |
Procedure:
Looking through the ocular with your right eye only (squint or cover your left eye), bring the specimen into focus by turning the coarse focus adjustment knob slowly until the specimen is generally in focus. Then turning the fine adjustment knob will bring the specimen (the coverslip edge) into sharper focus.
Question 23. Draw a sketch of several red blood cells observed in each of the three environments. Label their plasma membranes. In the last row write whether water moved into or out of the cells, or whether there was no change.
Hypertonic | Isotonic: | Hypotonic | |
Drawing: | Draw your picture by hand, take a picture, and paste it here. | Draw your picture by hand, take a picture, and paste it here. | Draw your picture by hand, take a picture, and paste it here. |
Description of relative size and shape of cells | |||
Water movement? |
Important: The red blood cells in the isotonic solution are in a “normal” osmotic condition, the same as when they were in the living animal. If we assume that the cell membrane is semipermeable, do you think water is moving in and out of the cell membrane in an isotonic environment? Describe the equilibrium.
Procedure:
Looking through the ocular with your right eye only (squint or cover your left eye), bring the specimen into focus by turning the coarse focus adjustment knob slowly until the specimen is generally in focus. Then turning the fine adjustment knob will bring the specimen into sharper focus.
Question 24. Notice these plant cells are shaped differently than the animal cells (human cheek cells). What gives these plant cells the brick-like shape?
Replace this text with your answer. |
Question 25. Note the green, jelly-bean shaped chloroplasts. Where are they located within the cell? In the center or around the edge near the cell membrane and cell wall?
Replace this text with your answer. |
Question 26. Sketch several Elodea (plant) cells in the space below to the left titled “Cells in tap water.” Label with a word(s) and an arrow pointing to their cell walls, central vacuoles, and chloroplasts.
Tap water | Hypertonic | |
Drawing: | Draw your picture by hand, take a picture, and paste it here. | Draw your picture by hand, take a picture, and paste it here. |
Description of relative size and shape of cells | ||
Water movement? |
Question 27. Where are the chloroplasts now in relation to the center or the cell wall?
Replace this text with your answer. |
Question 28. What had to have happened to the water in the central vacuole for the chloroplasts to be located where they are now after placing hypertonic solution on the slide?
Replace this text with your answer. |
Question 29. Observe the large plastic model of a plant cell. Notice that the central vacuole occupies most of the plant cell’s mass. The fluid concentration within this vacuole, called cell sap, contains a high concentration of salt, sugar and protein molecules. How does this help explain the large size of the central vacuole in the Elodea cells when placed in fresh tap water? (Hint: osmosis plays a part here.)
Replace this text with your answer. |
Question 30. The plant phenomenon observed above is called plasmolysis. In one word, what has happened to the Elodea cell?
Replace this text with your answer. |
Question 31. Notice that a plant cell, unlike an animal cell, has not only a plasma (cell) membrane, but also has a cell wall encasing it. Did the cell wall appear to swell, shrink or remain the same during your observations?
Replace this text with your answer. |
Question 32. Placing the wilted elodea cells (or wilted lettuce leaves that you might put into a salad) into tap water causes the plant cells to become turgid once again. Explain, in terms of osmosis, why the elodea leaf (or lettuce leaf) becomes “crisper” after placing it in tap water:
Replace this text with your answer. |
Question 33. Explain why the Elodea (plant) cells did not lyse when placed in tap water as did the animal cells (red blood cells). In other words, what is the reason that there is a difference between animal and cell’s response to being placed in these various solutions?
Replace this text with your answer. |
Clean up.
Returning your Compound Microscope to the Cabinet:
Clean up of the remainder of the lab materials.
Question 34. Compare animal and plant cells in varying solutions:
An animal cell placed in a(n)
Will
|
Causing the cell to
|
|
Hypertonic solution | ||
Isotonic solution | ||
Hypotonic solution |
A plant cell placed in a(n)
Will
|
Causing the cell to
|
|
Hypertonic solution | ||
Isotonic solution
(see your textbook) |
||
Hypotonic (tap water) solution |
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