Surface Area to Volume Ratio and the Relation to the Rate of Diffusion Aim and Background This is an experiment to examine how the Surface Area / Volume Ratio affects the rate of diffusion and how this relates to the size and shape of living organisms. The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen
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Surface area / Volume ratio Experiment Introduction: The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into
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Effects of the Ratio of Surface Area to Volume on Transformation Efficiency in Escherichia coli K-12 Eliana Pouchard‚ Steve Teng‚ and Cameron Wong Department of Microbiology & Immunology‚ UBC In the process of transformation‚ bact eria take up DNA fr om the environment through their cell wall. To induce competence in the cells‚ the DNA for uptake must first attach to the cell surface prio r to passing throug h the membrane. Previous studies on the effect of surface area on transformation
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does varying the surface area to volume ratio (2.0046: 1‚ 1.4923: 1‚ 0.9425: 1‚ 0.6480:1‚ 0.5970:1) affect the amount of heat lost over a period of 6 minutes of 50cm3 water with a temperature above 50? b. Prediction and Hypothesis. Make a prediction about what you expect to be the outcome. Explain your prediction using scientific ideas. I predict that the bigger the surface area to volume ratio‚ the lesser the heat will be lost. And the smaller the surface area to volume ratio the more heat
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AP Biology II 11/22/13 Title: The Effects of Surface Area to Volume ratio in Agar Introduction: What is an efficient way to maximize mass but minimize diffusion time in cell? Answer: An efficient way to maximize mass but minimize diffusion time in a cell is to increase its surface area. If you increase the surface area of a cell relatively to its volume‚ then the diffusion time will decrease. Materials: Agar cubes‚ bromothymol blue- pH indicator‚ vinegar‚ ruler‚ spatula‚ beaker
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[Type text] [Type text] [Type text] _An experiment on the effect of surface area to volume ratio on the rate of osmosis of Solanum tuberosum L._ BACKGROUND A cell needs to perform diffusion in order to survive. Substances‚ including water‚ ions‚ and molecules that are required for cellular activities‚ can enter and leave cells by a passive process such as diffusion. Diffusion is random movement of molecules in a net direction from a region of higher concentration to a region of lower concentration
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increase‚ which means that more substances have to be taken in and to also be removed. This is where the surface area to volume ratio comes into place; the reason why this ratio is so important is because the surface area of a cell essentially affects the rate of the transferring of useful substances (through diffusion and osmosis etc.) in and out of the organism. On the other hand‚ the total volume of the organism also affects the rate of the making of material inside the cell and the ability to hold
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Background information: 1) What is diffusion? Movement of a particular type of molecule from an area of high concentration to an area of low concentration. 2) How is diffusion used by living cells? Living cells bring in food‚ water and oxygen‚ and excrete wastes through the process of diffusion 3) List two body systems in vertebrates that are dependent on diffusion Digestive system and respirational system 4) What is meant by the term metabolism the chemical processes that occur within a
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Mathematics Volume of Solids Formulae for Volume of Solids Cube | Cuboid | Triangular Prism | Cylinder | Cone | Pyramid | Sphere | AnyPrism | s3 | lwh | ½bhl | Πr2h | 1/3πr2h | 1/3Ah | 4/3πr3 | Ah | A = area of the base of the figure s = length of a side of the figure l = length of the figure w = width of the figure h = height of the figure π = 22/7 or 3.14 1. Compute the volume of a cube with side 7cm. Volume of cube: s3 s = 7cm s3 = (7cm x 7cm x 7cm) = 343cm3 2. Compute
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diffusion. Diffusion is the process that cells use to obtain oxygen‚ water and food. Also‚ how they lose waste substances‚ for example‚ urea and carbon dioxide. Basically‚ Diffusion is when particles move from an area of high concentration to an area of low concentration. The surface area to volume ratio of the cell is an important factor in diffusion. It is the effect of this factor that will be investigated in this practical. Materials used in the practical consist of blocks of agar jelly containing
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