In recombinant DNA tech, To select a recombinant host from non-recombinants, we have different types of techniques available.
The one I'm talking about is the color selection method with the use of 2 markers(one antibiotic resistant gene for eg. Ampicillin, and the other Z-gene). We use Z-gene as the reporter gene. We use vectorless E. Coli as the host cell here.
Now, when we add add Ampicillin to the petri dish containing all of the variants (transformants and non-transformants), - the non-transformants will die. The transformants survive.
Now, we add X- gal to the replica of this petri dish, and we see some of the colonies turning blue due to the action of the enzyme β- galactosidase, which in turn is produced by the intact Z-gene in the plasmid. That means our blue colonies are the non-recombinant ones.
My question is-
The medium taken in the petri dish should contain glucose too , shouldn't it? Otherwise the recombinants would too give the colour reaction (X-gal is a homolog of lactose) due to the Z-gene present in their chromosomal DNA (operon concept).
The question maybe silly since it's kind of obvious we have to supply basic necessities for the growth of bacteria(which obioviously is glucose).
Just wanted to confirm in case we don't have glucose available ! =p
Answer
The Z gene you are talking about should be lacZ that encodes the $\beta-$galactosidase, which is made from $\alpha$ and $\omega$ peptides. Neither peptide is functional by itself.
$\beta-$galactosidase will cleave the glycosidic bond in X-gal and form galactose and 5-bromo-4-chloro-3-hydroxyindole. The latter product then dimerizes and oxidizes to 5,5'-dibromo-4,4'-dichloro-indigo, an intense blue product that is easy to identify and quantify.
Next, you should check out the genotype of your bacterial strain because they typically have only the $\omega$-peptide encoded in its genome. Thus, the enzymatic activity of the $\beta-$galactosidase is only recovered when the $\alpha$-peptide is in the environment, and the process is called $\alpha$-complementation.
Two commonly used strains for cloning and expression in E. coli are Dh5$\alpha$ and TOP10.
The genotype for Dh5a is like this:
F– Φ80lacZΔM15 Δ(lacZYA-argF) U169 recA1 endA1 hsdR17 (rK–, mK+) phoA supE44 λ– thi-1 gyrA96 relA1
Here, lacZΔM15 indicates that the M15 segment is lost from lacZ, and the mutant protein will only have the $\omega$-peptide.
The genotype for TOP10 is like this:
F– mcrA Δ(mrr-hsdRMS-mcrBC) Φ80lacZΔM15 ΔlacX74 recA1 araD139 Δ(ara leu) 7697 galU galK rpsL (StrR) endA1 nupG
Again, you will notice lacZΔM15, which is still a non-functional peptide segment after all.
Finally, to your question. Even if the there is no glucose as primary energy source in the medium and only X-gal is present, the operon of your host bacteria(I suppose it's E. coli) cannot produce the functional $\beta-$galactosidase on its own. Moreover, the recombinants, with the vector sequence encoding the $\alpha$-peptide disrupted, will not exhibit $\alpha$-complementation. No color reaction then! :D
Reference:
1.Genotypes of Invitrogen™ competent cells - TOP10, retrieved from: https://www.thermofisher.com/cn/zh/home/life-science/cloning/competent-cells-for-transformation/chemically-competent/top10f-genotypes.html.
2.Genotypes of Invitrogen™ competent cells - DH5a, retrieved from: https://www.thermofisher.com/us/en/home/life-science/cloning/competent-cells-for-transformation/chemically-competent/dh5alpha-genotypes.html.
3. Langley, K. E.; Villarejo, M. R.; Fowler, A. V.; Zamenhof, P. J.; Zabin, I. (1975). "Molecular basis of beta-galactosidase alpha-complementation". Proceedings of the National Academy of Sciences of the United States of America. 72 (4): 1254–1257. doi:10.1073/pnas.72.4.1254. PMC 432510 Freely accessible. PMID 1093175. 4.
No comments:
Post a Comment