2 Lab 2 Membrane Transport_Diffusion Dialysis Osmosis_Online_Sp21

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Lab Exercise 2 | Virtual Woo Sp 2021 1 Laboratory Exercise 2 Diffusion, Dialysis, and Osmosis Online Read the entire document. Complete all of the exercises. Fill in everything in RED. You may use the draw feature on Word or other to handwrite the information in the lab but you must type your answers to the lab assignment questions at the end of this document Activities: 1. Test the effect of temperature on diffusion: Video 2. Test the effect of molecular size on diffusion: Video 3. Test dialysis and osmosis in a dialysis bag: Video Lab Objectives: After today’s exercise you should be able to: 1. Explain the relationship between rate of diffusion and (a) temperature and (b) molecular weight of a solute; 2. Calculate molecular weight from a periodic table of elements ( see inner back cover of your textbook by Silverthorn for reference ); 3. Explain how dialysis of solutes across a semi-permeable membrane is related to size exclusion (“permeability”); 4. Explain how osmosis of water across a semi-permeable membrane is caused by a differential concentration of non-penetrating solutes (NPS) on either side of the membrane; and if osmosis occurs, the movement of water will cause a change in the weight of the dialysis bag. 5. Relate the results of the dialysis experiment to the means by which a living cell must regulate water balance between ICF and ECF; 6. Interpret the results of Benedicts, Biuret, and Lugols tests and the molecules they detect. 7. Explore the Scientific Method some more: a. What is the purpose of a hypothesis? b. Why is it important to limit variables in experimental situations? c. What is the purpose of including a negative control (and a positive control in some cases) in an experiment? Key Terms: (make sure you can define these terms): Diffusion, osmosis, gradient, cyclosis (cytoplasmic streaming), molarity, % solution, solution, solute, solvent, tonicity, osmolarity, dialysis, selectively permeable, passive transport. Background: All particles (atoms and molecules) are in constant, random motion and tend to scatter or disperse themselves equally from where they are densely concentrated to where they are less densely concentrated. This is the fundamental law of diffusion . The movement of molecules by diffusion can be a slow process. The rate of diffusion is governed in part by: ● Temperature ● Concentration gradient (the difference in the concentration between 2 regions),

Lab Exercise 2 | Virtual Woo Sp 2021 2 ● Size of the diffusing particles, ● Shape of the particles, ● Medium in which the diffusion occurs (solid/liquid/gas) The human body is mostly made up of aqueous solutions that are separated into ECF and ICF by selectively permeable membranes. In a living organism the rate of diffusion into and out of cells is increased by the action of streaming cytoplasm (cyclosis). You can see this streaming in this video of plant cells. The term “solution” will be used several times in this lab exercise. A solution is a homogeneous molecular mixture. Examples of solutions are sugar and water or salt and water. The water is called the solvent , and the sugar and the salt are called the solutes . For more information on diffusion, osmosis and the effects of the plasma membrane on the selective passage of solutes, refer to your textbook (Silverthorn, 8 th edition – pp.124-131) Here is a virtual lab experiment on “Diffusion, Osmosis & Tonicity” – Thistle tube experiment) . A power point presentation of this experiment is also available on the Canvas website for your course. I. Simple Diffusion: How temperature influences the rate of diffusion in a solution Experimental Question: Does temperature have an effect on the rate of diffusion? State your hypothesis: I hypothesis that diffusion will occur faster as temperature increases. Experimental design: A drop of the methylene blue stain is added to a beaker of cold, room temperature or hot water. The time it takes for the dye to become evenly distributed throughout the beakers will be recorded. What is the independent variable (IV) and what is the dependent variable (DV) in this experiment? The independent variable is the different temperatures that you are using to evaluate their effect on diffusion and the dependent variable is the rate of diffusion measured by time. Please watch this video testing the effects of temperature on rate of diffusion. You may be asked to recall the details of this experiment on the exam. Results: Table 1. Results of video demonstration illustrating influence of temperature on diffusion rate. Hot water * (_67__ ° C) Room Temperature water * (_23__ ° C) Cool water * (6___ ° C) * Time to reach diffusion equilibrium (minutes) 1 minute 40 minutes 3 hours *enter the temperature values and the times to even distribution given in the video Experimental Conclusion: An increase in temperature will increase / decrease the rate of diffusion along a concentration gradient. (circle the correct response) PART II. Effect of Molecular Weight (mass) on The Rate of Diffusion: Virtual Rates of diffusion can be very predictable if information about the diffusing substances is accurate and the medium of diffusion is held constant. The molecular weight of a substance can be used to estimate its size, molecules with greater molecular weights are larger in size . In this experiment you will be assessing the diffusion of three dyes which have three very different molecular weights: potassium

Lab Exercise 2 | Virtual Woo Sp 2021 3 permanganate, methyl orange, and congo red. You will use these dyes to investigate the relationship between the size of a molecule and rate of diffusion . Calculating Molecular Weight and Molarity : [ Before continuing, read through Figure 2.7 in the 7 th & 8 th editions of the textbook which discusses molecular mass and concentration of solutions]. To calculate the molecular weight (mass) of each of the dyes, add up the weights of the individual atomic components of each molecular formula. The molecular formula tells you how many (subscripted) of each element are in the whole molecule. No subscript means just one of that atom. For example, a molecule of water (H 2 O) contains two hydrogen atoms and one oxygen atom. By referring to the periodic table (see inner back cover of your textbook), you will find that hydrogen has an atomic weight of 1 gram per mole. The atomic weight of the element is (usually) the value under the symbol for the element. A mole is a constant number that is equivalent to 6.02 x 10 23 atoms or molecules of any atom or molecule. 1 mole of atoms or molecules will have a mass in grams that is equivalent to the atomic or molecular weight of the element or molecule. Oxygen has an atomic weight of 16 grams per mole; therefore, the total weight of water (H 2 O) is 1+1+16=18 grams per mole. Use the periodic chart below to calculate the molecular weight of the following molecules. NaCl = 23+35.5= 58.5 grams per mole C 2 H 4 O = 12+12+1+1+1+1+16= 44 grams per mole HCO3 - = 1+12+16+16+16 = 61 grams per mole Before watching the video calculate the molecular weights of the three dyes: (use your periodic table inside the back cover of the textbook and round the molecular mass to the nearest whole number) a. Potassium Permanganate: KMnO 4 = _ 39+54.9+(16*4)= 158 _ b. Congo Red: C 32 H 22 N 6 O 6 S 2 Na 2 = (12*32)+(1*22)+(14*6)+(16*6)+(32*2)+(23*2) = 696 c. Methyl Orange: C 14 H 14 N 3 NaO 3 S = (12*14)+(1*14)+(14*3)+23+(16*3)+32 = 327 Which of the 3 dyes has the greatest mass (mol. wt.)? Congo Red Which of the 3 dyes has the lowest mass (mol. wt.)? Potassium Permanganate

Lab Exercise 2 | Virtual Woo Sp 2021 4 In this experiment you will observe and compare the rates of diffusion of the three dyes through a gelatinous medium called agar (a semi-solid material made from seaweed similar to Jell-O). The dyes are dispensed on the top of the agar and after 2 hours the distance that the dye has diffused is measured with a metric ruler. Since the rate of diffusion is inversely related to the square root of the molecular weight of a substance, the heavier the molecule, the slower it will diffuse per unit time. With this in mind, state a hypothesis (in the space below) that will predict the relative rates of diffusion of the 3 dyes when placed in a gel-like substance called agar: (DO THIS BEFORE WATCHING THE VIDEO) HYPOTHESIS: I think the Potassium Permangante will diffuse the fastest followed by the methyl orange and the congo red will take the longest of the three dyes to diffuse. Please read the steps of the experiment below as you watch this video on the effect of molecular weight on diffusion. You may be asked about the experiment on the exam. Procedure: 1. 3 small test tubes filled with agar are placed in a test tube rack. Remove any water from the tops of the agar tubes by gently pouring water of the tube onto a kimwipe or tissue. 2. Take a set of 3 dyes with color-coded droppers. Label each agar tube with the code for one of the 3 dyes: PP=Potassium permanganate, MO=Methyl Orange, CR = Congo red . 3. Shake the dye bottles well and carefully place 150 µl of the specific liquid dye in each appropriately labeled tube without touching the agar. It will look like this: 4. Allow the tubes to sit undisturbed for 2 hours. 5. After 2 hours, measure the distance in millimeters (mm) that each dye has diffused into the agar. You may want to use a piece of white paper to hold tubes against, while measuring. 6. From the video, record the results in the table below . RESULTS Table 1. (title your table) :__Rate of Diffusion Dye name Molecular Mass Distance Diffused (mm) Rate of Diffusion (mm/hour) PP 158g/mole 12mm 6mm/hour MO 327 g/mole 8mm 4mm/hour

Lab Exercise 2 | Virtual Woo Sp 2021 5 CR 696g/mole 5mm 2.5mm/hour Follow up Questions on Agar-Dye Diffusion Rates (to be done after viewing the video): 1. Which of the dyes diffused the farthest (fastest)? Potassium Permanganate 2. Which diffused the shortest distance (slowest)? Congo Red 3. Do your results support your hypothesis? YES. If not, what may have caused the unexpected results? 4. What factor was the independent variable in this experiment? The molecular mass of dye 5. What variables were controlled in this experiment? Temperature, humidity, density of agar, volume of dye Graph your results in the space below; don’t forget to title your graph and label the X and Y axis with the Independent Variable and Dependent/measured Variable) Figure 1: Rate of Diffusion MW (g/mole) Distance (mm)

Lab Exercise 2 | Virtual Woo Sp 2021 6 Describe the relationship between the two variables shown on the graph. As the molecular mass increases the rate of diffusion goes down which means it takes longer for a molecule with a high molecular weight to diffuse than a molecule with a small molecular weight. Some Math to do: 1. How many grams of congo red are in 1 mole of this dye? 696g/mole 2. How many moles of congo red are in 1392 grams of this dye? 2 moles 3. How many moles of congo red are in 13.92 grams of this dye? .02 moles PART III. Diffusion of solvent (OSMOSIS) and Diffusion of solutes (DIALYSIS) across a semi- permeable membrane: VIRTUAL A membrane, whether it is a cellophane dialysis membrane or the membrane of a living cell, may be thought of as having small holes or pores. Some substances, if they are small enough, can pass through these membrane pores. Others, whose molecular sizes are too large, cannot pass through these holes. Non-living membranes that allow some molecules to pass while preventing others, due to size limitations, are said to be semi-permeable . Plasma membranes surrounding live cells contain transport molecules and electrically charged channels which, in addition to pore size, affect the ability of some molecules to pass, regardless of size; therefore, plasma membranes are said to be selectively permeable . It was noted in the introduction of this exercise that molecules are all in a constant state of random motion. Picture, in your mind, these molecules randomly bouncing against the side of a semi-permeable membrane (semi-permeable membranes have tiny holes in them). If the molecule that strikes the membrane is smaller than the holes in the membrane, the molecule will simply pass through the membrane. However, if the molecule is larger than the holes in the membrane, the free passage of the molecule will be blocked and the molecule will remain “trapped” on one side of the membrane. This is the concept behind the use of semi permeable dialysis tubing (dialysis tubing can be obtained with different pore sizes) to separate different molecules based on their size . The size of the pores in dialysis tubing determines which solutes may cross the membrane. The diffusion of any solute across a semi-permeable membrane is called dialysis . Within the human body, the greatest amount of dialysis occurs in the kidneys. The blood is dialyzed within the nephron capillaries. Water and small solutes are forced out of the blood through pores in the capillaries and empty into the nephron tubules. Some solutes are then reabsorbed into the blood by dialysis while water (the biological solvent) is reabsorbed by osmosis (a form of diffusion) (Osmosis is the diffusion of water across a membrane). Dialysis and osmosis do not require any energy expenditure by the cell; they are forms of passive transport . Furthermore, they do not usually require any carrier proteins to aid their movement across the membrane. You may want to view this brief 47 second video that demonstrates osmosis across a membrane: http://www.youtube.com/watch?v=sdiJtDRJQEc Concentrations of solutions can be calculated in several different formats, and they are very important in understanding physiology. The concentrations of the contents of your dialysis bag and dish solution in the

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