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)have constitutively active lacZ and lacY genes.
B)cannot activate either the lacZ or lacY gene.
C)can activate the lacY gene but cannot active the lacZ gene.
D)cannot metabolize lactose.
E)cannot transport lactose into the cell.

Question

Quizplus · Accepted Answer

The yellow colony indicates the presence of active β-galactosidase, which is produced by the lacZ gene. Since there is no lactose in the media, the only explanation for this color change is that the lacZ gene is constitutively active in the cells of the yellow colony. The lacY gene encodes for lactose permease, which is responsible for transporting lactose into the cell. There is no indication that this gene is involved in the color change.

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