Table 1. Results from Luria–Delbrück fluctuation tests.

	am432			am696			am1067
Fluctuation testa	1	2	3	1	2	3	1	2	3
$N_{\text{i}}$ (pfu)b	24.1	24.7	24.9	13.0	63.3	12.0	34.6	17.1	39.0
Initial revertants (pfu)c	0.00	0.27	0.53	0.00	0.27	0.00	7.2	0.40	0.80
$\overline{N_{\text{f}}}$ × 10−3 (pfu)d	8.39	5.69	6.47	14.3	11.6	3.19	11.4	1.63	6.21
$N_{\text{fH}}$ × 10−3 (pfu)e	2.18	2.25	4.28	4.85	6.15	2.67	4.69	1.18	4.07
Cultures with no revertants	9	21	12	9	6	18	13	20	11
Cultures with 1 revertant	7	3	6	5	6	4	6	3	7
Cultures with 2 revertants	3	0	3	4	6	2	1	1	4
Cultures with 3 revertants	1	0	2	2	1	0	0	0	0
Cultures with 4–100 revertants	3	0	1	4	5	0	1	0	2
Cultures with >100 revertants	1	0	0	0	0	0	3	0	0
P0f	9/23	21/24	12/24	9/24	6/24	18/24	13/21	20/24	11/24
Reversion rateg m × 104	1.1	0.24	1.1	0.69	1.2	0.90	0.42	1.1	1.3
Reversion rateh $\overline{m}$ × 104	4.4	0.60	1.6	2.0	2.3	1.1	1.0	1.6	1.9
Reversion ratei m(z) × 104	1.4	0.29	1.3	0.86	1.5	1.1	0.53	1.4	1.6
Reversion ratej m(z, R) × 104	—	—	—	—	—	—	0.34	0.91	1.0

^aEach test consisted of 24 parallel cultures.

^bAs determined from three independent titrations of the inoculum.

^cExpected initial number of revertants in 24 cultures as determined from three independent titrations of the inoculum.

^dArithmetic mean of $N_{\text{f}}$ obtained from 6/24 random cultures.

^eHarmonic mean of $N_{\text{f}}$ obtained from the same random cultures.

^fFraction of cultures showing no revertants, excluding cultures with >100 revertants from the total.

^g$m=-ln(P_0)/(\overline{N_{\text{f}}}-N_{\text{i}})$.

^h$\overline{m}=-ln(P_0)/(N_{\text{fH}}-N_{\text{i}})$.

ⁱCorrection for incomplete plating using a numerically estimated $P_0={\displaystyle {\sum }_k{Q}_k{(1-z)}^k}$, where $Q_k=P\left(k|m\left(N_{\text{f}}/z^{\prime }-N_i\right)\right)$.

^jCorrection for incomplete plating and selection, using a numerically estimated $P_0={\displaystyle {\sum }_k{Q}_k{(1-z)}^k}$, where $Q_k={\displaystyle {\sum }_{l=1}^{k}P\left(l|m\left(N_{\text{f}}/z^{\prime }-N_{\text{i}}\right)\right)P\left(k-l|R\right)}$. R was quantified only for am1067 and its CAG revertant by fitness assays. For am432 and am696 selection bias should be lower, because probably the fittest revertants were those carrying the wild-type CAG codon.