The 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
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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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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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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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[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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size and surface area to volume ratio. The experiment for cell size and diffusion was set to see how and how much water can go to the cells. This movement of water is called Osmosis. Osmosis is the movement of water molecules from an area of low concentration (lots of water) to an area of high concentration (little water) through a semi permeable membrane‚ demonstrated in ‘figure 1’. A semi permeable membrane is a membrane that only lets selected molecules to pass through it. In a plant water is taken
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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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In Parts A and C‚ the relationship between surface area and volume was investigated. Plasticine was formed into a cube and a sphere; both shapes were cut in half. It was found that plasticine volume should not vary‚ two halves have a greater surface area than a whole‚ and cubes have a greater surface area than spheres of the same volume. In Part B‚ the relationship between diffusion and surface area to volume ratio was investigated. Three agar-phenolphthalein-sodium hydroxide cubes of different sizes
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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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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 need to absorb their food‚ water and oxygen through their cell membrane and depending on the size of the cell membrane the rate at which it absorbs
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