**Transformation Experiment in E. coli**In a transformation experiment, DNA is collected from an *E. coli* donor strain of genotype cys(+) leu(-) thr(+) and used to transform a recipient of genotype cys(-) leu(+) thr(-). Initially, the treated recipient population is plated on a minimal medium supplemented with threonine. Many colonies are obtained. What are the possible genotypes of these colonies?**Options:**A. cys(-) leu(+) thr(+)B. cys(+) leu(+) thr(-)C. All are cys(+) and either + or - for leu and thrD. All are thr(+) and either + or - for cys and leuE. All are thr(-) cys(-) and either + or - for cysThis exercise presents a scenario in a microbiology experiment involving genetic transformation to determine possible genotypic outcomes of microbial colonies under supplemented growth conditions.

There are three modes of transfer of genetic information in bacteria. They are transduction, transformation, and conjugation. Conjugation is the transfer of DNA from one bacterium to the other. Transduction is the transfer of DNA with the help of a bacteriophage. Transformation is the direct uptake of naked DNA from the surroundings. Option A is not correct. The colonies formed won't be cys- as the cys- colonies cannot synthesize cysteine, and the medium doesn't contain cysteine. Option C is not correct. The colonies that are leu- cannot synthesize leucine, and the medium is not provided with leucine. Option D is not correct. The plate is supplemented with a minimum medium of threonine. Thus, the bacteria can survive even in the absence of threonine, but leu- and cys- cannot survive. Option E is not correct. Colonies that are cys- cannot survive on the plate as the plate doesn't have cysteine. Since the plate has threonine, any colony, whether it is thr+ or thr-, can grow in the plate. But since the plate is not provided with leucine and cysteine, the colonies have to produce these amino acids to survive in the plate. Thus, option B is a possible answer. The colonies can be cys+ leu+ and thr-. Another possibility is for the colonies to be cys+ leu+ and thr+. Thus, the correct answer is option B.