Transformation efficiency

The double deletion of the DH5α and KRX strain, reuslting in the phenotypes KRX Δalr ΔdadX and DH5α Δalr ΔdadX respectivly. This leads to a strict D-alanine auxotrophy, which can be complemented by a plasmid carrying one of the alanine racemases or D-alanine supplementation in the media.
The plasmid BBa_K1465401 used for the complementation carries the konstitutive alanine racemase alr and the chloramphenicol acetyltransferase for the chloramphenicol-resistance (Cm). Besides the plasmid contains RFP BBa_J04450 as a reporter for the characterization of the successful complementation.
For the proof of the antibiotic-free selection the plasmid was transformed in different concentrations into electrocompetent cells of DH5α Δalr ΔdadX. As it turned out, that the cell can store D-alanine for some time the incubation of the cells after the transformation was performed in normal SOC-Media without any supplemntation and in SOC-Media with supplemtation of 3 mM D-alanine to maintian the cell viability of the transformans, before they are able to express the alanine racemase essential for the accumulation of D-alanine and bacterial growth. After the incubation the cells were streaked out either of LB plates for the antibiotica-free selection via complementation of the D-alanine auxotrophy or were streaked out of LB plates containing 30 mg/L Chlormaphenicol as control. The successful transformation could be verified by the red color of the reporter RFP, while false-positive colonies remain white.
An example for the transformation of 0,33 ng of the plasmid BBa_K1465401 into the E. coli strain DH5α Δalr ΔdadX is shown in Figure 3. In the upper row different volumes from 50 µl, 100 µl and 200 µl were plated on LB containing Chloramphenicol and in the lower row the same volumes were streaked out only on normal LB for the selection without any antibiotic. Surprisingly it becomes obvious that the antibiotic-free selection is more efficient than the selection via chloramphenicol, because more colonies could be observed on each plate with an equal volume plated.

Figure 3: Colony forming units (cfu) by the selection of Chloramphenicol (upper row) and antibiotic-free selection (lower row). 0,33 ng of the plasmid BBa_K1465401 were transformed into the E. coli strain DH5α Δalr ΔdadX, incubated in SOC-Media supplemented with 3 mM D-alanine and streaked out in different volumes from 50 µl (left), 100 µl (middle) and 200 µl (right).

In Figure 3 and 4 the colony forming units (cfu) of the transformation of 0,52 ng plasmid BBa_K1465401 after one hour of incubation in either normal SOC-medium or supplemented with 3 mM D-alanine. First of all it can be seen, that the incubation of the transformation is possible also without the supplementation of D-alanine (blue bars), but there are not as much colonies compared to the incubation with 3 mM D-alanine (red bars). Besides the graphics show, that in the absent of D-alanine there are more false-positive phenotypic white colonies (cyan bars) compared to the supplementation of D-alanine (purple bar). In some cases this leads also to huge contaminations so that the amount of false-positive is even higher than the amount of correct colonies. This effect can clearly be seen by streaking out 200 µl of the transformants who were incubated in SOC-Media without D-alanine.
The most interesting aspect turns out by the comparision between the chloramphenicol selection and the antibiotica-free selection (AB-free), because it can be clearly seen, that the amount of positive red colonies is 2,88 +-0,45 (3 mM D-alanine supplemtation) and 5,91 +- 1,53 respectivly (without D-alanine) fold higher compared to the selection with the antibiotic chloramphenicol. This might be due to the active inhibtion of Chlormaphenicol, while the lack of D-alanine can be compensated by intracellular storage of D-alanine or eventually by and temporary grow arrest, until the alanine racemase is expressed.

Figure 4: Colony forming units (cfu) by the selection of Chloramphenicol. 0,52 ng of the plasmid BBa_K1465401 were transformed into the E. coli strain DH5α Δalr ΔdadX and incubated in normal SOC-Media or SOC-Media supplemented with 3 mM D-alanine. After one hour different volumes streaked out on LB containing 30 mg/L Chloramphenicol (n = 2 x 2).

Figure 5: Colony forming units (cfu) by the antibiotic-free selection of LB. 0,52 ng of the plasmid BBa_K1465401 were transformed into the E. coli strain DH5α Δalr ΔdadX and incubated in normal SOC-Media or SOC-Media supplemented with 3 mM D-alanine. After one hour different volumes streaked out on normal LB-Media (n = 2 x 2).

Another important aspect of an antibiotic-free selection approach is the quotient of false-positive and partically the portion of revertants that could also reduce this quotient in a antibiotic-free selection attempt. Therefore the portion of false-positive identified by its white color to the positive red colored transformants are shown in Figure 6. Besides the false-positive transformants in the antibiotic-free selection, there could also be observed false-positive using chloramphenicol selection. This false-positive could be bacteria with a spontane chloramphenicol-resistant, contaminations by bacteria with a Chloramphenicol-resistant or bacteria carrying a mutation within the expression of RFP (BBa_J04450) leading also to a white phenotype. Within the antibiotic-free selection an average portion of 2,83 % +- 0,09 false-positive have been identified, while the antibiotic-selection via chloramphenicol shows only an average of 1,47 % +- 0,77 false-positive. The slightly higher rate of false-positive of the antibiotic-free selection might due to some revertants which are able to accumulate D-alanine also without any alanine racemase by mutation and side reaction of another enzyme, but because of the higher transformation efficiency this effect is actually negligible.

Figure 6: Portion of false-positive colonies of classical antibiotic-selection using chlormaphenicol (red bars) and the antibiotical-free approach (orange bars). The false-positve colonies have been identified by their white phenotype. For the antiobiotic selection an average of 1,47 % +- 0,77 false-positive were estimated, while the portion amount to 2,83 % +- 0,09 for the antibiotic-free selection system.

To obtain more reliable data, the plasmid BBa_K1465401 were transformed in different concentrations (0,33 ng; 0,52 ng and 0,65 ng) and transformation attempt into the d-alanine auxotrophic E. coli strain DH5α Δalr ΔdadX and streaked out again in various volumes (50 µl; 100 µl and 200 µl) to estimate a valide transformations efficiency by using equation (1):
Transformation effiency [cfu/ng] = colony forming units / (Volume plated out [µl]/total volume [µl])*Amount of Plasmid-DNA [ng]
The estimated transformation efficiency for the transforamtion of 0,52 ng of BBa_K1465401 for the distinct volumes are shown in Figure 7, while the average for the transformation of different amounts of DNA are shwon in Figure 8. And again it becomes clear, that the antibiotic-free selection is a lot more efficient then the selection by the antibiotic chloramphenicol due to the inhibition of the antibiotic. And the other aspect obvious is that the transformation efficiency is reduced without the supplemtation of D-alanine due to the lytic effect when D-alanine is lacking and the crosslinkage of the peptidoglycane can not be performed by the cell. But even thought it is still possible.

Figure 7: Comparision of the Transformation efficiency for the transforamtion of 0,52 ng BBa_K1465401. The incubation was perfomed either in normal SOC media or in SOC-media supplemented with 3 mM D-alanine and the colony forming units were counted on LB_Cm as well as normal LB for the antibiotic-free selection (n = 2 x 2).

Figure 8: Comparision of the Transformation efficiency for different amount of the plasmid BBa_K1465401. The incubation was perfomed either in normal SOC media or in SOC-media supplemented with 3 mM D-alanine and the colony forming units were counted on LB_Cm as well as normal LB for the antibiotic-free selection (n = 2 x 2).