In this video we are going to take a look
at the cell surface area to volume (SA:V)
ratio and the limitations this has on cell
size.
The volume of the cell consists of the internal
regions of the cell.
Inside the cell there are a multitude of metabolic
reactions occurring which require the acquisition
of gases and chemicals as well as producing
waste.
The surface area of the cell is determined
by the cell membrane.
This cell membrane is responsible for regulating
the movement of molecules into and out of
the cell.
Gases and other molecules have to move through
the cell membrane into the cell and waste
products have to move through the cell membrane
to be eliminated from the cell.
Surface area, volume, and the SA:V ratio can
be easily calculated using a cube as a model.
Let’s look at a cube with a side length
of 1cm.
The volume of this cube is going to be all
the 3-dimensional space the cube takes up.
To work out the volume of a cube you times
the height by the width by the length.
So for this cube, it would be 1cm x 1cm x
1cm.
Which is the same as saying the volume is
1cm cubed (1cm3).
The surface area of this cube is the total
area that all the faces of the cube make.
If we were to deconstruct the cube, we can
see it is made of 6 sides, the area of each
can be calculated by times-ing the height
by the width, so 1cm x 1cm or 1cm2.
To get the area of the whole cube you would
then times this by the number of sides.
There are 6 sides, so it would be 6 times
1cm2 giving a surface area of 6cm squared.
To calculate the SA:V ratio you divide the
surface area by the volume.
The volume for this cube was 1cm3 and the
surface area was 6cm2
So for this cube we would divide 6cm2 by 1cm3
which equals 6cm-1.
The SA:V ratio gets smaller the bigger the
cube gets.
If the cube had a side length of 2cm, the
volume would be 2cm*2cm*2cm which equals 8cm3.
The surface area would be 2 cm x 2 cm by the
6 sides which equals 24cm2.
The surface area would be 24cm2 divided by
8cm3 which equals 3cm-1 which is smaller than
the surface area for the smaller cube.
If instead of a bigger cube, there were 8
smaller 1 cm length cubes.
The volume would be the same as the larger
cube, 1cm3 times 8 cubes equals 8cm3 total
volume for all the smaller cubes.
The surface area would be higher though.
6Cm2*8 = 48cm2 total surface area for the
smaller cubes compared to the 24cm2 for the
larger single cube.
This gives us a higher SA:V ratio also; 48cm2/8cm3
= 6cm-1, which is the same ratio as you get
with one single smaller cube.
Cubes are often used as a model for looking
at the surface area and volume of a cell.
The benefit of using cubes as a model is that
they are easy to visualise, measure and manipulate,
whereas real cells are hard to visualise,
measure and manipulate due to their microscopic
size as well as coming in different shapes.
The limitation of using cubes as a model is
that cells are mostly not cubic.
The size of the cell affects the SA:V ratio.
As seen in the model of the cubes, as the
cell gets bigger, the volume increases inside,
the surface area also gets bigger, but the
SA:V ratio gets smaller.
Cells are often limited in size by this SA:V
ratio.
When the cell is small it has a larger SA:V
ratio.
It has a smaller volume so needs less molecules
to be transported into the cell through the
membrane for metabolic processes and there
is also less waste to be removed through the
cell membrane.
At the same time, there is more cell membrane
relative to volume for this movement across
the membrane to occur.
In smaller cells, the molecules have to diffuse
through a shorter distance within the cell.
When the cells are larger, they have a smaller
SA:V ratio.
The volume is larger and has a higher need
for molecular nutrients to move across the
membrane into the cell, and more need for
the removal of waste across the cell membrane.
At the same time there is less cell membrane
relative to the volume for this to occur,
and in the cell, these molecules have a larger
distance in which to diffuse.
This limits the cells, if the metabolic rate
exceeds the rate of the exchange of the materials
needed in metabolism and the removal of waste,
the cell will in time die.
This is why growing cells often divide via
mitosis or binary fission to remain small
and be able to have a high enough SA:V ratio
to enable survival of the cell.
Cells are varied in shape, some are adapted
to increase the surface area and the SA:V
ratio.
One type of adaptation of a cell to maximise
the SA:V ratio is by forming long extensions
which increase the surface area of the cell
and in turn the SA:V ratio.
This is seen in nerve cells, where the cell
has many thin extensions branching out from
the cell body called dendrites.
A second type of adaptation of a cell to maximise
the SA:V ratio is by being thin and flat,
as this increases the surface area to volume
ratio.
This is seen in red blood cells, which are
flattened into a biconcave shape.
A third type of adaptation of a cell to maximise
the SA:V ratio is by having bristle-like extensions
called microvilli as they increase the total
membrane surface of the cell increasing the
SA:V ratio.
This is seen in cells in the intestine.
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