Question

Wild type E. coli cells are growing in a normal medium with glucose. They are transferred to a medium containing only lactose as the sugar. Which one of the following changes take place?
(a) The lac operon is repressed.
(b) All operons are induced.
(c) E. coli cells stop dividing.
(d) The lac operon is induced.

Answer 1

Hint: The Lac Operon is associated with the interaction of E. coli bacteria which grows in the normal medium and induces a change that causes the bacterium to digest something that is not its favorite food.
Complete solution:
The operon is a group of cells that is regulated together as a group and not individually. The lac operon is composed of three cells which are the lac Y, the lacZ and lac A. The E. coli is a bacterium whose favorite food is glucose. It is not that the bacteria feels it is yummy but because it has to spend less energy to metabolise it as it is simpler than lactose and fructose.
So, in case there is an environment in which glucose is present, the bacteria will first hunt down all the glucose and gather energy. It will jump to lactose only when the glucose concentration is null in the environment. It will break down and metabolise lactose just like it can metabolise glucose, however it will require more energy due to the complex nature of lactose. This also requires the induction of lac operon, which makes option D the correct answer.

So the correct answer is option d) The lac operon is induced.

Note:
In case lactose has disappeared from the environment, the time required for the cells to return to their normal state is determined by the effects of cell division and other regulatory processes. As the synthesis of lactose permease and β-galactosidase is stopped, their concentration decreases and the proteins around will dilute leading to the repression of the lac operon.