Surface area to volume ratio diffusion experiment

An important concept to understand for the year 11 is diffusion. It wasn’t until I performed the surface area to volume ratio diffusion experiment that I understood the idea of diffusion. Essentially, if you imagine having a group of students packed in the corner of a classroom, give it time and they will spread out to evenly fill the space. That’s all diffusion is. It is that simple!

Learning outcomes:

  • Understand what osmosis and diffusion are
  • Make predictions based on SA:V ratio and justify why a single cell can only grow so big
  • Know and understand the Surface area to volume ratio diffusion experiment

The surface area to volume ratio (SA:V) is one of the underlying themes throughout biology. It explains why cells are as small as they are, why we developed systems to deliver food and oxygen to our cells, why some animals are large and thin whilst others are fat and small.

The process of diffusion is the movement of solute from an area of high concentration to an area of low concentration. Passive diffusion occurs when a solute follows it’s concentration gradient hence energy is not required to move the solute.

Using this video we can visualise how diffusion works. Think of this as the water molecules outside the jelly trying to run inside the jelly and invade it. This is because there isn’t much water inside the jelly compared to outside.

Let’s first look at the surface area to volume ratio on the cellular level. Essentially, as an object gets larger it’s volume increases more then it’s surface area. Let’s use the example of a cube. It’s volume is it’s height * length * base but it’s surface area is it’s height * length * 6 (since there are six sides to a cube).

Graph for the Surface area to volume ratio diffusion experiment
The red line is the volume increasing whilst the blue line is the surface area increasing. This shows that their is an optimum ratio of SA:V and that as radius increases the volume increases at a greater rate to the surface area for this cubic model.

A cells surface area is a critical component in the exchange of material into and out of the cell. In essence, diffusion will occur at a greater rate if the surface area is larger. However, as the volume of a cell increases so does it’s requirements for nutrients and energy. This means the cell’s size must find the optimum between the amount of stuff a cell requires (dependent upon its volume) and the amount of stuff which can enter and leave the cell (dependent upon its surface area). For this reason a single celled organism can only be a certain size.

An interesting case of the SA:V ratio is the flat worm, it has an incredibly thin body keeping it’s volume low whilst have a large surface area due to it’s flat body. As a result it receives sufficient nutrients and expends waste at a rate which allows it t support its volume.

For multicellular organisms they have developed philological and structural adaptations which makes it possible for all cells to receive sufficient nutrients to support life. For example, the human blood vessels provide a stream of nutrient rich blood to most cells in the human body whilst carrying nutrients away – it would be impossible for oxygen to diffuse into our body through our skin and support all our cells. Hence we developed our lungs to allow us to dissolve the oxygen into our blood and our arteries carry it to our cells.

The surface area to volume ratio can also be represented by dividing the surface area by the volume to numerically represent the ratio. This means the larger the number the larger the surface area is to the volume.

This is also applicable to the concept of heat exchange. A larger surface area to volume ratio allows for more heat exchange then a smaller surface area to volume ratio. hence in colder climates it is more favorable to be smaller and fatter and in hotter climates its more favorable to be skinny and tall to maximize surface area to volume ratio.

Questions for the surface area to volume ratio diffusion experiment

Question 1

The surface area of a sphere is given by SA= 4*pi*r^(2), where as, the volume is given by V=(4/3) *pi *r ^(3). Calculate the Surface area and volume for a sphere with radius 1, 2 and 3 mm. Plot these results on a graph and use that to explain why a cell can only be so big. [6 marks]

Question 2

“The SA:V ratio rule only applies to single cellular organisms, since humans are multicellular organisms our SA:V ratio doesn’t matter”

Evaluate the validity of the students statement and explain why organisms such as humans can be so large. [8 marks]

Question 3

To minimize heat loss in cold climates, certain animals have developed adaptations to maximize their surface area to volume ratio. With the example of the wombat explain what this adaptation is. [3 marks]

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