SME
Tissue Engineering · Problem type
Seeding efficiency and growth
Use this for “Of 10⁶ cells seeded, 7.5 × 10⁵ attach. How many are there after 3 days at a 24 h doubling time?”
Diagram
75% of 1 × 10⁶ is 7.5 × 10⁵.
This is an example. Type your own values to start a new problem.
E
%
=
Nₐ
÷
Nₛ
×
100%
n
=
t (days)
÷
Td (h)
N
=
Nₐ
×
2
The bar shows the seeding only: then n doublings, each taking Td, fill the culture time t, and N is the count after them.
Formulas
E
=
100
×
Nₐ
Nₛ
75
=
100
×
7.5 ×
10
5
1 ×
10
6
n
=
t
Td
3
=
72
24
N
=
Nₐ
×
2
n
6 ×
10
6
=
7.5 ×
10
5
×
2
3
Formulas are worked in h. The step-by-step shows the conversions.
Assumptions
•
Seeding efficiency is the share of the seeded cells that attach: the rest wash off.
•
The attached cells grow exponentially with no lag or death, so N = Nₐ × 2ⁿ.
•
There are 24 hours in a day, so 3 days at a 24 h doubling time is 3 doublings.
Step-by-step
WE KNOW
Nₛ = 1 × 10⁶, Nₐ = 7.5 × 10⁵, Td = 24 h, t = 3 days
FIND
seeding efficiency (E), doublings (n), later cell count (N)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to h).
t = 3 days = 72 h (1 day = 24 h)
Step 2 · Find seeding efficiency (E)
Use E = 100 × Nₐ ÷ Nₛ
The attached cells as a share of the seeded cells, times 100 for a percent.
E
=
100
×
7.5 ×
10
5
10
6
E
=
7.5 ×
10
7
10
6
E
=
75%
Step 3 · Find doublings (n)
Use n = t ÷ Td
Count the doubling times that fit in the culture time.
n
=
72
24
n
=
3
Step 4 · Find later cell count (N)
Use N = Nₐ × 2ⁿ
Each
doubling
time
doubles
the
cells,
n
times
over.
N
=
7.5 ×
10
5
×
2
3
N
=
(7.5
×
10
5
)
×
8
N
=
6 ×
10
6
Check
Put the numbers back in. Both sides should match.
75
=
100
×
7.5 ×
10
5
10
6
✓
3
=
72
24
✓
6 ×
10
6
=
7.5 ×
10
5
×
2
3
✓
Topic check
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