SME
Microbiology · Problem type
Mutation rate from a fluctuation test
Use this for “20 of 50 cultures of 2 × 10⁸ cells have no mutants. What is the mutation rate?”
Chart
20 + 30 = 50.
This is an example. Type your own values to start a new problem.
C
Cultures
example
C₀
Cultures with no mutants
example
C₁
Cultures with mutants
worked out (not typed)
P₀
Share with no mutants
calculated
m
Mutations per culture
calculated
N
Cells per culture
example
μ
Mutation rate per cell division
calculated
Formulas
C₁
=
C
−
C₀
30 = 50 − 20
P₀
=
C₀
C
0.4
=
20
50
m
=
−ln(
P₀
)
0.9163 = −ln(0.4)
μ
=
m
N
4.58 ×
10
-9
=
0.9163
2.00 ×
10
8
Assumptions
•
Mutations arise at random while each culture grows, so the mutations per culture follow a Poisson distribution.
•
The share of cultures with no mutants is then P₀ = e−m, where m is the mean mutations per culture.
•
Growing to N cells takes about N cell divisions, so the rate per division is μ = m ÷ N.
Step-by-step
WE KNOW
C = 50, C₀ = 20, N = 2.00 × 10⁸
FIND
cultures with mutants (C₁), share with no mutants (P₀), mutations per culture (m), mutation rate per cell division (μ)
Step 1 · Find cultures with mutants (C₁)
Use C₁ = C − C₀
Every other culture grew at least one mutant.
C₁
=
50
−
20
C₁
=
30
Step 2 · Find share with no mutants (P₀)
Use P₀ = C₀ ÷ C
The share of the cultures with no mutants.
P₀
=
20
50
P₀
=
0.4
Step 3 · Find mutations per culture (m)
Use m = −ln(P₀)
The share with none is the Poisson chance of no mutations: take its natural log and change the sign.
m
=
−ln(0.4)
m
=
0.9163
Step 4 · Find mutation rate per cell division (μ)
Use μ = m ÷ N
Mutations per culture over the divisions in it.
μ
=
0.9163
2.00 ×
10
8
μ
=
4.58 ×
10
-9
Check
Put the numbers back in. Both sides should match.
30 = 50 − 20 ✓
0.4
=
20
50
✓
0.9163 = −ln(0.4) ✓
4.58 ×
10
-9
=
0.9163
2.00 ×
10
8
✓
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