Why does the surface area to volume ratio increase as the cell size decreases?

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Payal Khullar | College Teacher | (Level 2) Associate Educator

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Surface Area to Volume ratio (SA:V), as the name suggests, is the total surface area per unit volume of something (in our case, a cell). As we know, surface area has units in square and volume has units in cube. Hence, Surface:Volume (or Surface/ Volume) is measured in an inverse unit of length.

SA:V =`(SA)/(V)=x^2/x^3=x^(-1)`

Before answering the question, it is important to first understand the relationship of the surface area to volume ratio of a cell and cell size. A cell is like a compartment where numerous chemical activities take place. Like any living thing, cells grow and increase in size. The higher the volume (or capacity) of the cell, the larger such chemical reactions can take place. For such reactions to take place, a cell will need chemical raw materials or nutrients, which can enter the cell from the outside environment via diffusion in the cell membrane (cell membrane is like a covering present all over the cell surface that separates cell interior from the surrounding environment). This implies that a higher surface area will permit larger exchange of substances to take place.

As the cell increases in size, the volume of the cell also increases and so does the surface area. However, the rate of increase of volume and that of surface area of the cell are not proportional. Surface area increases in squares, while volume shows increase in cubes. In simple terms, surface area expands at a lower rate as compared to the volume. This leads to a low surface area to volume ratio when a cell increases in size.

In the same way, if the cell size is decreased, its volume and surface area will also decrease, but at unequal rates. Again, the volume would decrease faster than the surface area, leading to an increase of the surface unit to volume ratio. Hence, as cell size decreases, the surface area to volume ratio increases. We can even add that big cells have smaller surface area to volume ratio in comparison to small cells, and vice versa.

It is important to study Surface Area to Volume ratio in cell biology. For instance, after a considerable increase in the size of the cell, this surface area becomes so low as compared to the volume that it gets inadequate to provide the cell its minimum metabolic requirements through the exchange of chemical substances or nutrients with the environment. Hence, cell size, at this point, gets limited and the cell then undergoes cell division.

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parama9000 | Student, Grade 11 | (Level 1) Valedictorian

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As the volume remains the same, there is less space for the glucose in the cell to occupy, and therefore, the base/surface area increases as the liquid has less vertical space, forcing the liquid to spread out more across the surface area.

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rachellopez | Student, Grade 12 | (Level 1) Valedictorian

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Cells have to be able to take up and eliminate substances quickly as well as distribute substances. Most of this occurs by using the process diffusion, the movement of a substance from high concentration to low concentration.

As surface area and volume increase, they don't increase proportionally. Eventually the cell will be to big for substances to travel fast enough. So, in order to take in all the nutrients the body needs in a good amount of time the ratio will increase.

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Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

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Its easy to understand this if we look at a real example. 

Lets say we have a cube with side length x . The surface area is 6x^2 and the area is x^3 

If we take the derivative of these two functions with respect to time (for rate of change)

we get 

that dS/dt = 12x 
and dA/dt = 3x^2 

Therefore, we can see that the area function increases at a faster rate than the surface area function as x (the cell size) increases. 

so as cell size gets smaller, the area gets smaller faster than the surface area. Therefore, the ratio increases. 

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science2014 | (Level 1) Valedictorian

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For a cube,
 
Surface Area = length x width x number of sides
Volume = length x width x height
 
For a sphere,
 
Surface area = 4 (pi) r^2
Volume = (4/3) (pi) r^3
 
Hence, as the dimension or size of a cell increases, the increase in surface area varies with the square of the increased dimension whereas the increase in volume varies with the cube of the increased dimension. This means that the increase is not proportional and the volume increases much more rapidly than the surface area. This is why the ratio of surface area to volume decreases as cell size increase.
 
There is an optimum value of surface area to volume ratio when exchange of nutrients and wastes by diffusion becomes inadequate in a cell.
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chrisyhsun | Student, College Freshman | (Level 1) Salutatorian

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Simply put, think about the surface area (a measurement of area) and volume (a measurement of volume, of course) in terms of their units. Area is represented as units squared; volume is represented as units cubed. It makes mathematical sense, then, that as the area of something (in this case a cell) increases, the volume increases by even more because rather than squaring the factor, you are cubing it. On the flip side, as the cell size decreases, the volume decreases more drastically than surface area. Therefore, the ratio of these two components - surface area to volume - will increase as cell size decreases since the surface area is only somewhat decreasing but the volume is dramatically decreasing.

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