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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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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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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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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Cell Biology- Osmosis‚ Cell Size and Diffusion and Enzymes 1.0 INTRODUCTION Cells are the basic building blocks of all living things. They provide structure for the body‚ take in nutrients from the food‚ convert those nutrients into energy‚ and carry out specialized functions. Cells also contain the body’s genetic material and can make copies of themselves. A cell is also a metabolic compartment where many different chemical reaction occur. There are two types of cells‚ eukaryotic and prokaryotic
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AREA (i) The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonal of the rhombus is 22 cm‚ find the length of the other diagonal. (ii) The floor of a rectangular hall has a perimeter 250m. If the cost of paining the four walls at the rate of Rs 10 per m2 is Rs 1500. Find the height of the hall. (iii) A room is half as long again as it is broad. The cost of carpeting the room at Rs
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Task 2C In this experiment I am investigating the relationship between surface area and the volume of a model cactus and to determine whether this relationship affects the loss of water. We used oasis blocks‚ an absorbant material used by florists as a substitute for a cactus‚ that would soak up the water‚ and thus ideally show the percentage water loss by weighing the before and after evaporation weights. Independent Variable The independent variable is the total surface area for each size of
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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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ITEM PAGE Background Research 3-4 Experiment Hypothesis 5 Variables 5 Equipment 5 Risk Assessment 5 Method 6 Results 6 Analysis/Discussion 7 Conclusion 8 Acknowledgments/Bibliography 8 Table of Contents Background Research As an object becomes bigger its surface area and volume increases but the surface area to volume ratio decreases‚ this is because volume increases quicker than the surface area; as volume is three dimensional. This concept applies to cells and reaction rate because cells
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