SME
Biomaterials
Degradation
Use this for “A PLGA scaffold starts at Mₙ = 100 kDa and hydrolyzes with k = 0.05 per day. What is Mₙ after 14 days, and what is its half-life?”
Chart
Mₙ(14) ≈ 49.66
Level
Mₙ₀
2
=
50
meets
the
curve
at
t
1
2
=
13.86
This is an example. Type your own values to start a new problem.
Mₙ (kDa)
=
Mₙ₀ (kDa)
e
−
day⁻¹
×
days
t½ (days)
=
ln
2
k (day⁻¹)
M½
Half the start
worked out (not typed) · kDa
Formulas
Mₙ
=
Mₙ₀
×
e
−
k
×
t
49.66
=
100
×
e
−0.05 × 14
t
1
2
=
ln(2)
k
13.86
=
ln(2)
0.05
M
1
2
=
Mₙ₀
2
50
=
100
2
Assumptions
•
Water reaches every bond, so the bulk hydrolyzes evenly: dMₙ/dt = −kMₙ, first order in the ester bonds.
•
The scaffold keeps its mass at first: it only loses mass once chains are short enough to dissolve.
•
k and t use the same time unit, days here.
Step-by-step
WE KNOW
Mₙ₀ = 100 kDa, k = 0.05 day⁻¹, t = 14 days
FIND
Mₙ at time t (Mₙ), half-life (t½), half the start (M½)
Step 1 · Find Mₙ at time t (Mₙ)
Use Mₙ = Mₙ₀ × e−k × t
The chains shorten at a rate in step with their length: Mₙ is Mₙ₀ times e raised to −kt.
Mₙ
=
100
×
e
−0.05 × 14
Mₙ
=
100
×
e
−0.7
Mₙ = 100 × 0.4966
Mₙ = 49.66 kDa
Step 2 · Find half-life (t½)
Use t½ = ln(2) ÷ k
At
t
1
2
the
chains
are
half
as
long,
so
kt
1
2
=
ln
2.
Divide
ln
2
by
k
.
t
1
2
=
ln(2)
0.05
t
1
2
=
0.6931
0.05
t
1
2
=
13.86
days
Step 3 · Find half the start (M½)
Use M½ = Mₙ₀ ÷ 2
The
level
the
curve
crosses
at
t
1
2
:
half
the
starting
chain
length.
M
1
2
=
100
2
M
1
2
=
50
kDa
Check
Put the numbers back in. Both sides should match.
49.66
=
100
×
e
−0.05 × 14
✓
13.86
=
ln(2)
0.05
✓
50
=
100
2
✓
Topic check
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