nce gene on the plasmid serves the purpose of a selectable marker that provides a new characteristic to the transformed cell not possessed by the non-transformed cell, thereby distinguishing from the non-transformant on a selective media.
The ‘No DNA’ control on the LB plate (which contains no antibiotic or X-gal) is expected to show lawn pattern of bacterial growth implying that the bacterial cells are viable and can grow in the absence of the antibiotic ampicillin. However the "No DNA" on the LB blue plate is expected to show no growth as they do not contain the necessary gene for the restriction enzyme to survive in the medium with ampicillin.
Tube 2 shows all blue colonies which illustrates that transformation of cells with plasmid pCK103 confers the bacterial cells with the ability to grow in ampicillin medium due to the presence of the resistance gene and also convert X-gal in the medium to produce blue colonies by the action of lac Z gene present in the plasmid which encodes - galactosidase involved in lactose metabolism.
3). x g of the plasmid pCK103 were added to the E. coli during the transformation. From your results calculate the total number of transformants produced by this amount of DNA. Remember that the 100l sample is only half of the total transformation and the 10l sample is, of course, a twentieth.
4). This value is known as the transformation frequency and is a measure of the efficiency of the process. It is to some extent dependent on the plasmid used but for pCK103 a value of 106 - 107 transformants per g may be expected under ideal conditions. How does your value compare to this? Why do you think it differs?
The value differs from the ideal transformation efficiency predicted for pCK103. A number of factors affect transformation efficiency such as the actual DNA concentration and amount of DNA, heat shock, length of time for expression after transformation as well as the selective plates used.