Consider two imaginary cells. Both are cube shaped (the same dimension in all directions) The first cell has a side length of 10 micrometers. The second has a side length of 100 micrometers. Calculate the surface areas and volumes of both of these cells then calculate the sa/v for each of them. The SA, V and Sa/V dived out of each of theses cells. which one of these would be most efficeint at feeding itself? Why? If we added a third cell with a side length of 50 micrometers how would its feeding efficiency compare to our first two cells ( no calculations needed) Why ?
Finally if our least efficient cell needed to increase its feeding effiency to match the most efficient cell what is one way it might do it? Oh - that's without actually changing its volume. Be specific about how the cell must change.
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The side length of the first cell = 10 micrometers.
Since the cell is of cubic shape, its surface area (SA) is given by 6a2 where a is the side length of the cube.
Therefore, the SA of the first cube = 6x(10x10-6 )2 (1 micrometer =10-6 meter )
Volume of the cell (V) = a3 = (10x10-6 )3 = 10-15 m3
for the first cell , Surface Area/Volume = SA/V =6x10-10 /10-15
= 6x105 per meter
Similarly, for the second cell whose side length = 100 micrometers,
Surface Area = 6x(100x10-6 )2 = 6x10-8 m2 ...
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