Question

X-gal is a synthetic molecule which is a substrate for beta-galactosidase (b-gal). When cells are grown on a medium containing X-gal, they turn blue if the cells are expressing b-gal. In the absence of b-gal expression, the cells are white. You are attempting to clone a gene of interest into a lacZ reporter gene of a plasmid and introduce your recombinant molecule into a bacterial host sensitive (i.e. not resistant) to ampicillin. The plasmid contains a gene for ampicillin resistance at a position other than the cloning site. You grow your potential transformants on media containing both ampicillin and X-gal. Which of the following results would most likely indicate a successful clone?
cells which form blue colonies on the ampicillin plates
cells which form white colonies on the ampicillin plates
cells which form light blue (half blue, half white) colonies on the ampicillin plates
cells which do not grow on the ampicillin plates
None of the above results would indicate a successful clone.

Answer 1

Answer:

The correct answer is "cells which form white colonies on the ampicillin plates".

Explanation:

X-gal staining, also known as blue–white screen, is a common approach to detect colonies that express the gene of interest that has been incorporated into its genome. The basis of this technique is introducing the gene of interest into a vector that interrupts the expression of beta-galactosidase, which means that in a medium with X-gal the cells that were successfully transformed will be white. In this case, the cells that are transformed will also be resistant to ampicillin. Therefore if they growth in a media with both ampicillin and X-gal, a successful clone will form white colonies.