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 the cell so the cell would probably not survive. Single celled organisms can survive as they have a large enough surface area to allow all the oxygen and nutrients they need to diffuse through. Larger multi celled organisms need specialist organs to respire such as lungs or gills.

1. A block of gelatin which has been dyed with universal indicator should be cut into blocks of the following sizes (mm). 5 x 5 x 5
10 x 10 x 10
15 x 15 x 15
20 x 20 x 20
10 x 10 x 2
10 x 10 x 10 (Triangle)
10 x 15 x 5
20 x 5 x 5
The triangle is of the following dimensions.
The rest of the blocks are just plain cubes or rectangular blocks. Universal indicator is a neutral indicator. In the alkali conditions of the gelatin it is blue and when it gets exposed to acid it turns red. Gelatin is used for these tests as it is permeable and so it acts like a cell. It is easy to cut into the required sizes and the hydrochloric acid can diffuse at an even rate through it. 2. A small beaker was filled with 100cm³ of 0.1 molar Hydrochloric acid. This is to ensure that all the block sizes are fully covered in acid when dropped into the beaker. 3. One of the blocks is dropped into this beaker and the time for the entire universal indicator to disappear is noted in a table...

...Introduction
Cells have to interact with their environment, chemicals and water and in order to do so they must be able to move across the cell membrane and the cell. The movements within a cell are called Diffusion. When molecules move across a cell membrane it is known as Osmosis. Diffusion is the process by which molecules of a substance move from areas of higher concentration of that substance to areas of lower concentration. Diffusion can be the transfer of anything anywhere. However, that is not true for osmosis. Osmosis is diffusion, but a specific type of diffusion. Osmosis is only the diffusion of water molecules through a selectively permeable membrane and a concentration gradient. Only certain molecules can cross the membrane to either go in our out of the cell.
If outside the cell there is a lower concentration of molecules than inside the cell, it is called hypotonic and water will move from outside the cell into the cell. If the solution outside the cell has a larger concentration of dissolved materials the solution is hypertonic, so the water would move from the cell into the solution.
Aim
The aim of this experiment was to verify the concept of Osmosis and Diffusion with a semi-permeable membrane (dialysis tubing), it will be exposed to different environments and concentration gradients.
Hypothesis
I thought that that dialysis tubing would end up weighing more as there would be less water and more...

...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 living organism in order to maintain life.
5) Why cant a single cell grow very large?
If it grows too large it will not be able to take sufficient food and oxygen or eliminate waste molecules fast enough.
6) Why must large organisms be multicellular?
To overcome the problem of small cell size. Metabolism is the total of all the chemical processes that take place in the body. These chemical processes convert the food you eat into the energy and materials needed for all life processes.
Inputs and Outputs
Anabolism and Catabolism
• Metabolic processes can be regarded as either anabolic or catabolic.
Catabolism
• Catabolic reactions are destructive metabolic processes during which complex substances are broken down into simpler ones.
• Catabolic processes release energy.
• Cellular respiration is a good example of a catabolic process. Below is a summary of the cellular respiration process.
glucose + oxygen —> water + carbon dioxide...

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_An experiment on the effect of surfacearea to volumeratio 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 order to reach equilibrium. Diffusion does not require any energy input. Diffusion is needed for basic cell functions - for example, in humans, cells obtain oxygen via diffusion from the alveoli of the lungs into the blood and in plants water is obtained through the root hairs via osmosis.[1: http://www.tiem.utk.edu/~gross/bioed/webmodules/diffusion.htm]
Osmosis is a special kind of diffusion that involves water molecules passing through a partially permeable membrane, in this case, the plasma membrane of cells. Cell membranes have tiny holes that allow small molecules to pass through but not larger ones. The rate of osmosis is dependent on many factors - temperature, pressure and surfacearea to volumeratio, which is being explored in this experiment.
AIM
The aim of this biology laboratory experiment is to explore the...

...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 the volume of a cuboid (also known as a rectangular prism) with the dimensions 4cm by 13cm by 9cm.
Volume of a cuboid: lwh
l = 4cm
w = 13cm
h = 9cm
lwh = (4cm x 13cm x 9cm) = 468cm3
3. Compute the volume of a triangular prism with a base length of 60cm, a base width of 8cm, and a height of 10cm.
Volume of a triangular prism: ½bhl
½b = (8cm x 1/2) = 4cm
h = 10cm
l = 60cm
½bhl = (4cm x 10cm x 60cm) = 2400cm3
4. Compute the volume of a cylinder which is 2m tall and has a radius 75cm. Convert this litres.
Volume of a cylinder: πr2h
π = 3.1415
r2 = (75cm)2 = 375 cm2
h = 2m = 200cm
πr2h = 235612. 5 cm3
cm L = 1cm 0.001 L
235612.5 cm3/ 1000 = 235.6125 L
5. Compute the volume of a cone with a radius of 200cm and a height of 0.75m.
Volume of a cone: 1/3πr2h
1/3π = 1.047
r2 =...

...Investigating Ratios of Areas and Volumes
In this portfolio, I will be investigating the ratios of the areas and volumes formed from a curve in the form y = xn between two arbitrary parameters x = a and x = b, such that a < b. This will be done by using integration to find the area under the curve or volume of revolution about an axis. The two areas that will be compared will be labeled ‘A’ and ‘B’ (see figure A). In order to prove or disprove my conjectures, several different values for n will be used, including irrational, real numbers (π, √2). In addition, the values for a and b will be altered to different values to prove or disprove my conjectures. In order to aid in the calculation, a TI-84 Plus calculator will be used, and Microsoft Excel and WolframAlpha (http://www.wolframalpha.com/) will be used to create and display graphs.
Figure A
1. In the first problem, region B is the area under the curve y = x2 and is bounded by x = 0, x = 1, and the x-axis. Region A is the region bounded by the curve, y = 0, y = 1, and the y-axis. In order to find the ratio of the two areas, I first had to calculate the areas of both regions, which is seen below. For region A, I integrated in relation to y, while for region B, I integrated in relation to x. Therefore, the two formulas that I used...

...SurfaceareaSurfacearea is the measure of how much exposed area a solid object has, expressed in square units. Mathematical description of the surfacearea is considerably more involved than the definition of arc length of a curve. For polyhedra (objects with flat polygonal faces) the surfacearea is the sum of the areas of its faces. Smoothsurfaces, such as a sphere, are assigned surfacearea using their representation as parametric surfaces. This definition of the surfacearea is based on methods of infinitesimal calculus and involves partial derivatives and double integration.
General definition of surfacearea was sought by Henri Lebesgue and Hermann Minkowski at the turn of the twentieth century. Their work led to the development of geometric measure theory which studies various notions of surfacearea for irregular objects of any dimension. An important example is the Minkowski content of a surface.
Definition of surfacearea
While areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of area requires a lot of care. Surfacearea...

...The larger the surfacearea, means there can be more “paths” from the sides of the body that are capable of releasing this heat particles, and reaching thermal equilibrium faster. This is what happens when a hotter body is subjected to a colder one.
Research Question:
How does the surfacearea to volumeratio affect heat loss in organisms?
Hypothesis:
I hypothesize that the larger thesurfacearea to volumeratio, the more heat will be lost and vice versa. In this experiment, there will be a series of sizes of round bottom flasks, each having a different surfacearea to volumeratio. The smallest round bottom flask will have the biggest surfacearea to volumeratio, therefore the heat loss in that particular flask will be most, whereas the biggest flask will have the least surfacearea to volumeratio, therefore heat loss will be the least.
The reason behind the difference in rate of heat loss can be explained by the fact that the surfacearea to volumeratio indicates the surfacearea, a particular unit volume has. This, in other words, states that...