Lab Cell Size

Cell Size Lab Introduction: Cells come in a variety of sizes. Some red blood cells are only 8 µm in diameter. Nerve cells can reach lengths up to 1 meter. Most living cells, however, are between 2 and 200 µm in diameter. Why can’t organisms be just one giant cell? 1. Diffusion limits cell size: cells require a constant supply of glucose and oxygen to carry out __________________________________ and to produce large amounts of _______________. These substances, and waste products such as carbon dioxide, move through the cytoplasm by diffusion. It has been estimated that it takes a molecule of oxygen only a fraction of a second to diffuse through the cytoplasm from the plasma membrane to the center of a typical cell that is 20 µm. What would happen if the cell got bigger? Although diffusion is efficient over short distances, it becomes slow and inefficient as the distance becomes larger. Hypothetically, a cell with a diameter of 20 cm would have to wait months before receiving molecules that enter the cell. Because of this time restriction the cell/organisms would die if they were one large cell. 2. A cells DNA limits size: The second reason why cells are small is because they usually only contain one nucleus. If a cell doesn’t have enough DNA to program its metabolism, it cannot live. When a cell is larger it requires more enzymes and parts to function correctly, therefore it needs more instruction. In many large cells there are more than one nuclei to ensure that all cell activities will be carried out quickly. 3. Surface area-to-volume ratio: The third reason why cells are limited in size is that as a cell’s size increases, its volume increases much faster than its surface area. Note: Surface area = length of a side X width of a side X number of sides Volume = length X width X height Surface area to volume ratio = surface area / volume The smallest cell has a volume of 1mm3 and a surface area of 6mm2. If the side of the cell is doubled to 2 mm, the surface area will increase fourfold. Calculate what happens when the cell side is 4mm. Volume _________________________ Surface area ____________________________________________. If cell size is doubled, the cell would require eight times more nutrients and would have eight times more waste to excrete. The surface area would only increase by a factor of 4. Therefore, the plasma membrane would not to have enough surface area through which oxygen, nutrients, and waste could diffuse. The cell would either starve to death or be poisoned by the build p of waste products. Objective: In this lab, the diffusion rate of iodine will be observed through 4 potato cubes of different size. The cubes will turn black when it comes in contact with iodine. Iodine will slowly diffuse into the potato cubes and change its color from clear to black. Procedure: 1. Obtain potatoes from your instructor. Using a knife cut the potato into cubes following the given measurements below. Plan out your cuts ahead of time, it is suggested that you cut out the large cube first. Only potatoes will be given to your group. Block A: Block B: Block C: Block D: 0.5 cm sides 1.0 cm sides 2.0 cm sides 3.0 cm sides 2. Place all four cubes in the bottom of a beaker so that they do not touch each other or in the sides of the beaker. 3. Pour in enough sodium hydroxide to cover all four cubes. Allow 20 minutes for the diffusion process. GENTLY stir the solution so that diffusion can take place evenly over each cube. 4. While the cubes are soaking, Calculate the surface area for each cube and record the answer in table 1. Surface area = length of a side X width of a side X number of sides Calculate the volume of each cube and record the answer in table 1. Volume = length X width X height Find and record the surface area to volume ratio for each cube in table 1. Surface area to volume ratio = surface area / volume 5. Remove the blocks from the solutions, rinse underwater, place on a paper towel and slice each cube in half. 6. Quickly measure to the nearest 0.1 cm across the area with no iodine on each cube (clear area) 7. Calculate the surface are, volume, and surface area to volume ratio for the portion of the cube with iodine (clear cube) and record it in table 2. Table 1: Cube Side length (cm) Surface area (cm2) Volume (cm3) Surface area to volume ratio (surface area : volume) A B C D Percent Volume of Cub That Received Iodine = Total Volume of Original Cube (cm3) (Total cube volume) – (volume of cube that has not changed color) (total cube volume) “Unchanged” Cube Side Length: Subtract depth of diffusion (both sides) from original cube dimensions (cm) Volume of unchanged cube (that has not changed color) (cm3) X 100% Percent of total volume of cube that received iodine (see formula above) Questions: 1. If each cube represented a living cell, and the iodine solution was a substance needed within the cell, what problem might the largest cell have? 2. Examine your data in table 2. What pattern do you notice in the relationship between cube size and the % of the cell that recieeved iodine? 3. Examine your data in table 1. Describe what happens to the surface area and the volume as the cell grows larger. 4. Still considering table 1, what happens to the ratio between surface area and volume as the cell grows larger? 5. According to your data, which cell was most successful at receiving the needed nutrient (iodine solution) in the allowed time? 6. What can you say about the surface area to volume ratio that will best meet the needs of living cells? 7. Based upon your data, explain why cells are so small.