SME
Principles of Biology I
Cell structure
Use this for “A spherical cell has a radius of 5 μm. What is its surface area to volume ratio, and what happens when it doubles in width?”
Diagram
A
=
4πr²,
V
=
4
3
πr³,
so
A
V
=
3
r
r
=
5
μm:
A
V
=
3
5
=
0.6
μm⁻¹
r
×
2
=
10
μm:
A
V
=
0.3
μm⁻¹,
1
2
as
much
membrane
for
each
μm³
This is an example. Type your own values to start a new problem.
A
÷
V
=
3
÷
r (μm)
=
A ÷ V (μm⁻¹)
A = 4πr² and V = 4/3 πr³, so A ÷ V = 4πr² ÷ (4/3 πr³), which cancels to 3 ÷ r.
A
Surface area
worked out (not typed) · μm²
V
Volume
worked out (not typed) · μm³
Formulas
A
=
4π
×
r
2
314.1593
=
4π
×
5
2
V
=
4
3
×
π
×
r
3
523.5988
=
4
3
×
π
×
5
3
A
V
=
3
r
0.6
=
3
5
Assumptions
•
The cell is a sphere of radius r.
•
Nutrients come in through the surface, but every μm³ inside uses them, so a bigger cell feeds each μm³ through less membrane.
•
The second cell is twice as wide, drawn at the same scale.
Step-by-step
WE KNOW
r = 5 μm
FIND
surface area (A), volume (V), surface area to volume (A ÷ V)
Step 1 · Find surface area (A)
Use A = 4π × r²
A sphere’s surface is 4π times the square of its radius.
A
=
4π
×
5
2
A
=
314.1593
μm²
Step 2 · Find volume (V)
Use V = 4/3 × π × r³
A
sphere’s
volume
is
4
3
π
times
the
cube
of
its
radius.
V
=
4
3
×
π
×
5
3
V
=
4
3
×
125
×
π
V
=
500
3
×
π
V
=
523.5988
μm³
Step 3 · Find surface area to volume (A ÷ V)
Use A ÷ V = 3 ÷ r
4π
r
2
4
3
πr³
cancels
to
3
r
:
the
ratio
falls
as
r
grows.
A
V
=
3
5
A
V
=
0.6
μm⁻¹
Check
Put the numbers back in. Both sides should match.
314.1593
=
4π
×
5
2
✓
523.5988
=
4
3
×
π
×
5
3
✓
0.6
=
3
5
✓
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Actual size from a micrograph
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