1 Experiment Membrane Transport Objectives ► Referring to energy, what two ways can substances enter a cell? What is active transport? What is passive transport? How is osmosis related to diffusion? How can we demonstrate active transport? How can we demonstrate Brownian movement? How can we demonstrate diffusion (2 ways)? How can we demonstrate osmosis (3 ways)? In terms of relationships between substances, how can we define “hypertonic”, “isotonic”, and “hypotonic”? What is the relationship between the size of a molecule and its rate of diffusion? ____________________________________________________________________________ Supplies ► Materials Needed Active Transport: Baker’s Yeast 0.75% Na2CO3 0.02% Neutral Red Erlenmeyer Flasks Flame or Hot Plate Brownian Movement: India Ink (Carmine Dye or Whole Milk may be substituted) Slides and Cover Slips Diffusion: Beaker of Distilled Water 1.5% Agar-agar (or Gelatin) in petri plate 2 Potassium Permanganate (KMnO4) Crystals Potassium Dichromate (K2Cr2O7) Crystals Methylene Blue Crystals Metric Ruler (to measure in mm) Artificial Cell: Dialysis Tubing or Sacs Sugar String Beakers Distilled Water Scale Osmosis and Red Blood Cells: Distilled Water; Salt Solutions: 12-15% Salt Salt Solutions: 0.85% Salt (This is human, isotonic saline.) Blood (Human or Animal): Lancets and Alcohol if Human Blood is used Slides and Cover Slips Microscope ___________________________________________________________________________ Key Terms ► Active Transport Brownian Movement Cell Physiology Dialysis Diffusion Gradient Hypertonic Hypotonic Isotonic Membrane Transport Osmosis Passive Transport ____________________________________________________________________________ Introduction ► The individual cell is a dynamic microcosm, demonstrating in miniature all the processes and events that occur in the macrocosm, making the apparent function of the whole organism the actual function of many individual cells working in unison. It is important that we understand how the individual cell, the microcosm, functions so that we can more fully understand how the organism as a whole, the macrocosm, function. For instance, we can say that if an action potential (or nerve impulse) is to be generated in a part of the nervous system, a certain electrical stimulation must be present and certain ions must be 3 moving through appropriate channels. What we sometimes forget is that these events—in this case the electrical stimulation and the ionic movement—occur at the cellular level. What seems to be happening in the organism is happening only because the events are occurring at the level of the individual cell. We could use similar analogies for every system in the body. In lecture you examined the molecular intricate of the phospholipid bilayer known as the cell membrane, and you became aware that the cell membrane is selectively permeable, meaning that only certain substances can enter and leave the cell by freely crossing the membrane. You know, for instance, that the membrane is replete with channels, gates, and carrier molecules that either facilitate, inhibit, or repel assorted ions and molecules as they randomly approach the demarcation barrier. This demarcation barrier, the cell membrane, is functional in maintaining cellular integrity. You also know that since the cell is microcosm, ions and molecules must cross the barrier, both as nutrients entering the cell and as wastes leaving the cell. In cell physiology, we examine how events occur within the cell. Cellular functions follow the basic principles of physiology. Many of these functions do not lend themselves to easy demonstration, particularly this early in an introductory course. However, at this point we can demonstrate a number of functions directly related to membrane transport. And membrane transport is one of the main keys to cell physiology when we study the nervous system and the digestive system.) ____________________________________________________________________________ Preparation ► I. A. 1. Background and Protocol Explanation of Terms Be certain you have a working knowledge of the following terms: Membrane Transport Active Transport Passive Transport Diffusion Osmosis Brownian Movement Gradient Membrane transport is any process, active or passive, by which a substance moves from one side of a membrane to the other side of the membrane. We have already established that the membrane acts as a barrier that can facilitate, inhibit, or repel substances, and yet certain substances must cross the barrier. Active transport requires cellular energy because the substance traversing the barrier either cannot do so without a push or cannot do so in sufficient quantity to maintain cellular integrity. Active transport is absolutely essential in maintaining the functioning organism. Active transport is accomplished via ionic pumps. Manu of our carrier systems work because of active transport. Gradients, which enable work to be done, are maintained because of active transport. Active transport can best be demonstrated by visualizing some very common events, both in the laboratory and in daily life. Passive transport requires no input of cellular energy. Several categories of passive transport can be identified. Nevertheless, the most common readily identifiable form of passive 4 transport is diffusion, the net movement of molecules from an area of higher concentration to an area of lower concentration. Osmosis is form of diffusion. In the human system, osmosis is the diffusion of water. Passive transport can be easily demonstrated by both diffusion and osmosis experiments. Brownian movement, while technically not diffusion, is covered here with the passive transport experiments because Brownian movement demonstrates molecular motion. It is precisely because of this innate movement that molecules are able to move passively from one point to the next point—including from one side of a membrane to the other side. A gradient is any difference in intensity between two sides of a demarcation. For instance, a concentration gradient would be a difference in concentrations of a substance on each side of a demarcation. A physical barrier may or may not be present. We can also have temperature gradients, electrical gradients, flow gradients, and pressure gradients. B. General Procedural Points Your instructor may set up parts of this experiment as a demonstration. If so, answer the questions from your observations. 1. Work with a partner in Part II. 2. Divide the work between you and your partner in Part III so that each of you is doing one of the experiments. 3. If you use human blood, READ THE CAUTIONS GIVEN IN Anatomy of the Blood AND Blood Testing Procedures. ____________________________________________________________________________ Experimentation ► II. Active Transport Active transport is the energy-requiring movement of a substance across a barrier or a gradient. This transport could not take place without the input of energy. In the biological world, energy is supplied by the living cell. We can observe the principles of active transport by examining whether or not the uptake of a substance will take place in living and/or nonliving cells. Neutral red is a dye that is actively transported into the living yeast cell. Neutral red is red in an acidic solution and yellow in a basic solution. A sodium carbonate (Na2CO3) solution is basic. A. Experiment #1: Active Transport 1. Obtain two Erlenmeyer flasks, 2 g yeast, 50 ml solution carbonate, and 50 ml neutral red. 2. Into Flask #1 place half the yeast and half the sodium carbonate. Mix well and heat gently. Boil the solution for 2 or 3 minutes. When the solution is cool, add about half the neutral red. The solution should be yellow. 5 3. To Flask #2 add the remaining Na2CO3 and neutral red. Gently swirl in the remaining yeast and watch for a color change. Concept Check 1 Jot down your observations and discuss with your partner the differences in color between Flask #1 and Flask #2. B. Additional Observations Tip: In Flask #2 either the dye entered the cell or some substance was given off by the cell to change the color to red. Which answer seems most logical? Why? How might you test the solutions to see exactly what happened? Concept Check 2 Discuss with your partner the involved in windmills and water pumps. What is the role of energy in these apparatuses? How can you relate these functions to the role of active transport in the cells? III. Passive Transport (Diffusion) Passive transport is the movement of a substance across a membrane or a gradient without the expenditure of cellular energy. We have several experiments that demonstrate passive transport. A. Experiment #2: Brownian Movement All molecules with a temperature above absolute zero are in a state of constant motion. Because of this motion, molecules collide regularly and randomly. The manifestation of these collisions is called Brownian movement. Some disagreement exists as to the exact nature of Brownian movement. Nevertheless, for our purposes, the exact results are the same and we can see Brownian movement whenever we have a liquid mixture composed of differently sized molecules. Brownian movement is not actually a part of diffusion, but if the molecules were not in motion, they could not diffuse across a membrane or a gradient. Thus, in order to understand diffusion we should gain some insight into molecular movement. India ink is a suspension of carbon in water, along with ethyl alcohol and acetone. (Carmine dye and whole milk, the alternate substances for this experiment, are also molecular suspensions.) 1. Obtain a slide, cover slip, and microscope. 2. Place a drop of India ink on a slide with a small drop of water. Cover carefully with a cover slip and allow the solution to rest for 2 or 3 minutes. 6 3. Observe the suspension under both low and high power. If possible, also observe the suspension under oil. Keep the light as low as possible so you manitian maximum contrast. If your suspension is too dark, you may wish to start over using a larger drop of water. 4. Sketch what you see. 5. Describe Brownian movement based on your observations. B. Experiment #3: Diffusion in Liquids Diffusion—the net movement of molecules from an area of higher concentration to an area of lower concentration—can be demonstrated in two different ways. 1. Obtain a beaker of water and a crystal of potassium permanganate (KMnO4). 2. Drop the crystal into the water and watch the diffusion process. Use Figure 1 to sketch the initial diffusion path. Note the time and observe the beaker at 5-minute intervals. Note the time when you think diffusion is complete. How long did it take? Table 1 Petri Plate Diffusion Time Time KMnO4 15 min 30 min 45 min 60 min 75 min 90 min ____ C. Experiment #4: Diffusion In Solids K2Cr2O7 Methylene Blue 7 1. Obtain an agar-agar or gelatine petri plate and ruler. Also obtain equal crystals of potassium permanganate (KMnO4), potassium dichromate (K2Cr2O7), and methylene blue. 2. Construct an imaginary equilateral triangle on the petri plate and put one crustal at each angle, per Figure 2. At 15-minute intervals measure the radius (in millimeters) that each substance has diffused from the original crystal mark. Use Table 1 to record these distances. Figure 2 Petri Plate. Concept Check 3 The molecular weight for KMnO4 is 158, for K2Cr2O7 is 294, and for methylene blue is 320. In general terms, what is the relationship between the distance traveled and the molecular weight of each crystal? [Challenge: Although your measurements are probably not exact enough to derive a mathematical formula, do you see that the diffusion rate of any substance might well be inversely proportional to the square root of the molecular weight?] D. Osmosis Overview Osmosis is the movement of water from an area of higher water concentration, across a selectively permeable membrane, to an area of lower water concentration. That is, from an area of lower solute concentration to an area of higher solute concentration. Movement is always toward equilibrium. Dilution is toward equilibrium. Dialysis is the movement of a nonwater particle across a barrier. This movement is based on the size of the particle in relation to the size of the pores within the barrier. In the living system this barrier is the cell membrane. We can perform three experiments to demonstrate the principles of osmosis and dialysis. Work with a partner. 8 Table 2 Artificial Cell Results ____ Concept Check 4 Explain your results by answering these questions: At what height did each solution stop rising? Then what happened? What differences did you notice between the sugar and salt osmometers? How do you explain those differences? If you had tasted the water at the end of the experiment, what would you have noticed? How would you explain this? 9 E. Experiment #6: Osmosis (Artificial Cell) This experiment is outlined in Figure 4 and Table 2 1. Tip: keep in mind the sizes of various ions, molecules, and membrane pores. Obtain 5 beakers, 5 artificial cells (dialysis tubing or sacs), string, sugar, scale. 2. Make 3 sugar solutions, one 40%, and one 10%. Set up 5 artificial cells. Into cell #1 put some of the 40% solution, into cell #2 the 20% solution, into cell #3 the 10% solution, and into cell #4 and cell #5 put distilled water. 3. Take the 5 beakers. Into beaker #1 and #2 put distilled water, into #3 put 10% sugar, into #4 put 20% sugar, and into #5 put 40% sugar. Weigh each artificial cell and place it into its corresponding beaker. 4. At the end of the laboratory period—or later in the day, depending in the directions from your instructor—remove the cells from the beakers, wipe the cells with a paper towel, and reweigh. Calculate the percent differences in weight for each cell. Chart you results on Table2. What type of differences do you see in the different cells? How would you explain these differences? Test condition #1 Sac—40% Sugar Beaker—Water #2 Sac—20% Sugar Beaker—Water #3 Sac—10% Sugar Beaker—10% Sugar #4 Sac—Water Beaker—20% Sugar #5 Sac—Water Beaker—40% Sugar Weight at start Weight at finish 5. Below is a partially constructed bar graph. Complete this bar graph to show the relative changes in the weights of the sacs. 6. Repeat #1-4 above using time as a variable. Use the space at the end of this experiment to construct a graph demonstrating your findings. F. Experiment #7: Osmosis (Red Blood Cells) 10 A solute is any substance dissolved in any solvent. In biology, the solute is usually ionic, molecular, or particulate. The solvent is generally water or water-based. Hypertonic, isotonic, and hypotonic are terms used to describe the relationship between the solute concentrations on two sides of a membrane or gradient. Keep in mind that these are relationship terms. We shouldn’t simply state, “Solution X is hypertonic.” To show the relationship, we should say, “Solution X is hypertonic to Y.” Hyper-means greater (more, larger), so a hypertonic solution has a higher particulate concentration than the substance with which it is being compared. Iso- means same, so the two isotonic solutions would be of the same concentration. Hypo- means beneath (less, below), so a hypotonic solution has a lower particulate concentration than the substance with which it is being compared. When particulate movement occurs between two concentrations, the net movement is always toward equilibrium. Note the words “net movement”. If pore size is adequate, some movement against equilibrium does occur because molecules are in constant motion. We are concerned here with net movement. We are also concerned here only with passive movement. (Occasionally, cell membrane integrity or active transport mechanisms may influence net movement. We are not considering those aspects of movement here.) Note: Diseases such as cholera disrupt osmotic equilibrium. The cholera toxin causes changes in the membrane of the cells of intestinal epithelium. These cells lose fluid to the lumen. Because the intestinal cells have become more hypertonic to their environment, water from the interstitium enters these cells. This water is also lost to the lumen. Because of the increased hypertonicity of the interstitium, water leaves adjoining cells and enters the interstitium and eventually the epithelial cells. Dehydration progresses rapidly. Diarrhea is a characteristic of cholera because of the massive water loss. The large intestine (which is basically unaffected by the cholera toxin) cannot reabsorb the water as rapidly as it is lost from the small intestine. I. PREPARATION OF A STOCK BLOOD SUSPENSION Equipment: ovine blood isotonic saline (0.15M NaCl; 0.9% NaCl is another way of expressing the same concentration) 1 box of parafilm 1 pipette (10cc) 1 pipette suction pump 1 test tube 1 Pasteur pipette Each of the experiments in this lab requires a few drops of a stock suspension of ovine red blood cells, prepared as follows: 1. Put 10 ml of isotonic saline solution (0.9 % NaCl) in a clean test tube. 2. Add about 10 drops of ovine blood. 3. Seal the opening of the test tube with parafilm. 4. Mix thoroughly by putting your finger over the end of the tube and inverting twice gently. 11 II. DEMONSTRATION OF OSMOSIS The effects of osmosis can be observed when red blood cells are exposed to salt solutions at concentrations different from that of blood plasma. Hemolysis or crenation results when water molecules enter or leave the cell. EXERCISE A Equipment: stock suspension of blood prepared in I. distilled water isotonic saline (0.15M NaCl)) 2 pipettes (10cc) 1 Pasteur pipette 1 pipette suction pump 2 test tubes 1. Add 5 drops of stock blood suspension to each of 2 test tubes containing the following solutions: i) 5 ml distilled water ii) 5 ml isotonic NaCl. 2. Observe the print on this page through each of the test tubes. The solution of isotonic saline and blood is turbid and the printed characters appear blurry: this indicates that there is no hemolysis of the red blood cells (the print seen through the solution in the test tube appears blurry because the red blood cells are intact and thus diffract the light). The solution of distilled water and blood is clear and the printed characters can be seen very clearly: this indicates that hemolysis of the red blood cells has occurred (the print seen through the solution in the test tube appears very clear because the red blood cells have exploded and thus do not diffract the light anymore). Note: Keep these tubes for future reference: they will be your control tubes for the exercises B & C. III. EFFECT OF MOLECULAR SIZE ON PERMEABILITY OF THE CELL MEMBRANE The cell membrane is differentially permeable. Ionized solutes such Cl--, in spite of their small molecular size, do not pass through rapidly. Moreover, active transport pumps for Na+ and Ca++ remove these ions from cytoplasm as fast as they enter cells. The cell membrane is more permeable to some of the unionized polar solutes used in this experiment. EXERCISE B Equipment: stock blood suspension distilled water 0.3 M urea 0.3 M ethylene glycol 0.3 M glycerol 0.3 M glucose 5 test tubes 5 pipettes 1 pipette suction pump 1 timer 12 Relevant properties and formulas for the solutes used in this experiment are given in Table 1. Note that molecular size (diameter) roughly parallels molecular weight. All of these compounds are polar since they have either -OH (hydroxyl) or -NH2 (amino) groups. 1. Put 5 ml of 0.3 M urea in a test tube. 2. Add 5 drops of red cell suspension and mix gently. 3. As the first drop of blood reaches the solution, start the timer (or note the exact time if you prefer using your watch). Determine the time required for hemolysis to occur. You do this by holding the tube in front of the print on this page; as hemolysis occurs, the print will appear more and more clear. The hemolysis time is the time required for the solution in the tube to go from turbid to clear (and the print to go from blurry to clear). You should stop your timer when the print seen through the test tube containing urea is as clear as the print seen through the control tube containing distilled water. 4. Repeat the procedure using ethylene glycol, glycerol and glucose. 5. Record hemolysis times in Table 2 and on the blackboard. Table 1: Properties and Formulae of Solutes and Solvents SUBSTANCE MOLECULAR WEI WEIGHT (g) Urea urea urea 3.6 MOLECULAR DIAMETER (A) FORMULA H2 N C=O H2 N ethylene glycol 62 ~3.6 CH2OH CH2OH glycerol 92 6.2 CH2OH CHOH CH2OH glucose 180 8.6 CHO CHOH CHOH CHOH CHOH CH2OH 13 Table 2: Your results. SUBSTANCE Urea Ethylene glycol glycerol glucose TIME TO HEMOLYSIS IV. EFFECT OF LIPID SOLUBILITY ON PERMEABILITY OF THE CELL MEMBRANE This experiment examines the relation between lipid solubility and cell membrane permeability. Recall the structure of the plasma membrane. Ions and water-soluble molecules must pass through pores in the membrane because they cannot dissolve in the central lipid layer of the membrane. On the other hand, lipid-soluble molecules can dissolve in the lipid portion of the membrane and diffuse across easily. Many organic molecules contain both non-polar bonds, which confer lipid solubility, and polar bonds, which confer water solubility. Hydroxyl (-OH) and amino (-NH2) groups promote water solubility by virtue of the polar bonds between hydrogen and oxygen (in -OH groups) or between hydrogen and nitrogen (in -NH2 groups). On the other hand, carbon-carbon or carbon-oxygen bonds are nonpolar. When molecules contain bonds of both types they are soluble (to a greater or lesser extent) both in water and in lipids. The more polar bonds present, the greater is the water solubility, and the less soluble the compound is in lipids. EXERCISE C Equipment: stock blood suspension 0.3 M glycerol 0.3 M triacetin 0.3 M diacetin 3 test tubes 3 pipettes 1 pipette suction pump 1 timer The partition coefficient of a compound is a measure of its relative solubility in water and lipids. Partition Coefficient = Solubility in Olive Oil / Solubility in Water As the partition coefficient increases, fat solubility increases and water solubility decreases. Note in Table 3 that the partition coefficient increases as the number of polar hydroxyl groups decreases. 14 Determine times to hemolysis for each solution, following the procedure outlined for EXERCISE B. (Repeat glycerol test - do not use previous results). Record hemolysis times in Table 4 and on the blackboard. Table 3: Partition coefficients and formulas. SUBSTANCE MOLECULAR WEIGHT (g) PARTITION COEFFICIENT FORMULA glycerol 92 0.00001 CH2OH │ CHOH │ CH2OH triacetin 218 0.36 diacetin 176 0.1 CH2‐O‐COCH3 │ CHO‐COCH3 │ CH2‐O‐COCH3 CH2-O-COCH3 │ CHOH │ CH2-O-COCH3 Table 4: Your results. SUBSTANCE glycerol Triacetin diacetin TIME TO HEMOLYSIS V. OBSERVATION OF OSMOSIS UNDER MICROSCOPE EXERCISE D Equipment: ovine blood distilled water isotonic saline (0.15M NaCl) 0.3 M NaCl 1 box of Kimwipes 3 microscope slides and cover slips 1 microscope 4 Pasteur pipettes masking tape 15 1. Label 3 microscope slides: "dist. water", "saline" and "0.3 M NaCl". 2. Label the 4 Pasteur pipettes: "dist. water", "saline", "0.3 M NaCl" and "blood". 3. Dilute the stock ovine blood 2:1 with isotonic saline (10 drops isotonic saline to 5 drops of blood). 4. Place a drop of isotonic saline on the appropriate microscope slide. Add a small drop of the diluted ovine blood on the top of the drop of isotonic saline. Add a cover slip and examine immediately under the microscope. In the center of the slide, the red blood cells are piled up several layers thick and thus it is very hard to determine the shape of individual red blood cells. Make sure that you look where the layer of blood is thinner and try to find a spot where you see individual cells clearly. Use high power and reduced light intensity. Sketch outlines of cells. 5. Repeat this procedure twice, but use distilled water and 0.3M NaCl instead of isotonic saline. CLEAN-UP See “At the beginning of the lab” on first page. Leave your space clean and tidy. 16 WORKSHEET Name: ID No.: Instructor: Course / Section: Partner’s Name (if applicable): Date: Additional Activities 1. Try any of the diffusion experiments again. This time using exact measurements. Compute the exact rate of diffusion for each substance. Use a variety of mathematical and/or statistical formulas to discover whether or not other relationships exist. 2. Design additional experiments to demonstrate movement across a membrane—either active or passive. 3. Refer to the experiment on red blood cell osmosis. What would you need to know if you wanted to determine the relative amount of water moving across the cell membrane? Answer to Selected Concept Check Questions 1. In flask #2 the neutral red must have been transported into the living yeast cell. In Flask #1 the yeast cells were killed so no transport could occur. 2. These pumps correspond to the sodium/potassium pump in the cell membrane. The sodium/potassium pump actively pumps three sodium ions out of the cell for every two potassium ions pumped into the cell, thus maintaining the membrane potential. You may wish to figure out other macroscopic situations where active energy input is required. 3. The potassium permanganate should have traveled about twice as far as the potassium chromate. The potassium chromate should have traveled just a bit further than the methylene blue. Name: General Questions Date: 17 Use your own paper if necessary. 1. Define active and passive movement across a cell membrane. 2. Explain what happened to the yeast solutions. Why was the solution in Flask #2 a different color than the solution in Flask #1? 3. What caused the Brownian movement you observed? 4. Explain what happened when you dropped the potassium permanganate (KMnO4) Crystals into the water. 5. If your diffusion rates do not match the expected diffusion rates, how might you explain the discrepancy? 6. In which of the experiments did osmosis occur? Dialysis? In which direction did these events occur? 7. Check your lecture text and find out what effective osmotic pressure is? How does it relate to cellular osmosis? 8. Explain hypertonic, isotonic, and hypotonic in terms of the red blood cell experiment. 9. List several factors (other than the ones we demonstrated by these experiments) that might influence the rate of either active or passive transport. 10. Why are sugar and salt used for food preservations?