Imagine that you've added the same indicator used by Monod to E. coli colonies growing on a plate with glucose and no lactose. One colony is yellow and the remainders are white. This result suggests that cells of the yellow colony
A) cannot metabolize lactose.
B) cannot transport lactose into the cell.
C) have constitutively active lacZ and lacY genes.
D) cannot activate either the lacZ or lacY gene.
E) can activate the lacY gene but cannot active the lacZ gene.

Question

Quizplus · Accepted Answer

The staining method used is based on the ability of bacteria to break down lactose into glucose and galactose, which is then metabolized into an intermediate that can react with a color indicator (X-gal) and produce a blue color. The fact that one colony is yellow indicates that it cannot break down lactose into glucose and galactose, but it is also producing a color intermediate. This suggests that the lacZ and lacY genes are constitutively active in this colony, meaning that they are being expressed regardless of the presence of lactose.

Imagine That You've Added the Same Indicator Used by Monod