Diffusion

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Rutgers University *

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Biology

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Jan 9, 2024

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Diffusion Estimated Completion time: 1.5 hours Website: http://lab.concord.org/embeddable.html#interactives/sam/diffusion/4-semipermeable.json Introduction: Define the following terms: Passive transport: is how a cell moves materials in and out without expending energy; no energy is required unlike active transport. Diffusion: substances use this method to travel short distances in organisms; some substances include oxygen, waste products, and food. The substances move from a higher concentration to a lower concentration. Semipermeable membrane: a membrane that permits solvent molecules through (water) but is too small to allow solute molecules through. Particle: extremely small constituent of matter; this may include atoms, molecules, or even ions. For this investigation you will examine the effect of the size of pores in semipermeable membranes on the rate of diffusion. Hypothesis: (Select one) A (small / medium / large ) pore size will allow diffusion to occur the fastest. Procedure: Use the Concord: Diffusion Across a Semipermeable Membrane simulator and examine the data that was collected. Then, calculate the averages and answer the follow up questions. Simulation 1 1. Set the pore size slider to approximately 25% (about the size of 1 large green particle) to simulate a small pore in a cell membrane and check the box for Trace Molecule. 2. Hit play and pause it when the traced molecule (with the dotted line following it) crosses the membrane into the smaller section on the left. 3. Record this time in the data table. 4. Repeat this process 2 more times for trial 2 and 3. 5. Calculate the average. Simulation 2 6. Repeat steps 1-5, this time with the slider at 50% (about the size of 2 large green particles) to simulate medium size pores. Simulation 3 7. Repeat steps 1-5, this time with the slider at 75% (about the size of 3 large green particles) to simulate large size pores. Data: Small Pore Size Trial # Diffusion time (sec) 1 9.6
2 22.2 3 14.1 Average 15.3 *Remember, the average = (the sum of all trials) divided by (the number of trials) Calculation: (9.6+14.1+22.2)/3 Medium Pore Size Trial # Diffusion time (sec) 1 9.2 2 3.1 3 2.0 Average 4.77 *Remember, the average = (the sum of all trials) divided by (the number of trials) Calculation: (9.2+3.1+2.0)/3 Large Pore Size Trial # Diffusion time (sec) 1 4.8 2 3.2 3 2.8 Average 3.6 *Remember, the average = (the sum of all trials) divided by (the number of trials) Calculation: (4.8+3.2+2.8)/3 Analysis: 8. After examining your data, what effect does the size of a pore have on the rate of diffusion? After examining my data, it becomes apparent that the size of a pore is directly proportional to the rate of diffusion. As the size of a pore increase, the rate of diffusion increases; vice versa. Since small pores are closely packed particle matrices, there isn’t much room for the particles to diffuse adequately. On the other hand, as the porosity of the membrane increases, the permeability increases, as well as the space for particles to diffuse. As a result, it reduces the rate of diffusion.
9. In a paragraph, explain what effect very small pores and their diffusion rate would have on the workings, mechanisms, and health of a real cell. Since small pores reduce the diffusion rate, cell migration is limited. In addition to this, the distribution of nutrients and removal of waste products can be limited. As a result, it can lead to necrotic regions and the death of tissue. However, a reduced diffused rate helps maintain the electrolyte balance in a cell. Small pores provide more adhesion support, although they reduce the cell growth space, hindering it. 10. In a paragraph, explain what effect very large pores and their diffusion rate would have on the workings, mechanisms, and health of a real cell. Since large pores allow for a faster diffusion rate, it could disturb the normal functionality of a cell. More specifically, many electrolytes and water molecules would propagate through the semipermeable membrane causing immense amounts of pressure on the cell membrane. Due to this osmotic process, the cell membrane the cell could rupture, leading to lysis. As mentioned for small pores, the distribution of nutrients is limited. On the other hand, large pores will decrease surface area, limiting cell attachment. Conclusion: Was your hypothesis correct? Why or why not? Yes, my hypothesis was correct. I assumed that a more porous cell membrane would allow more particles to diffuse since there’s more available space. According to the data above, as the porosity of the cell membrane increases, the time it takes for a selected particle to diffuse decreases.
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