Passage III
In the human gut, E. coli bacteria break down lactose (milk sugar) into 2 sugars: glucose and galactose. Th reaction is catalyzed by the enzyme β-galactosidase.
The E coli gene lacZ contains information needed to synthesize β-galactosidase. lacZ is located within the lac operon (a portion of DNA). The lac operon also contains the lac operon operator (LOO), a DNA region that regulates lacZ (see Figure1).
If LOO is active, lacZ is active, so β-galactosidase is synthesized. If LOO is inactive, lacZ is inactive, so β-galactosidase is not synthesized. The activity of LOO is regulated by protein 1 (P1), a protein found in E. coli cells.
Three students propose models to describe the regulation of LOO by P1 in E. coli.
Student 1
If lactose is absent from a cell, P1 binds to LOO, inactivating LOO. If lactose is present, it binds to P1, causing a conformational change (shape change) in P1. As a result, P1 cannot bind to LOO. When P1 is not bound to LOO, LOO is active (see Figure 2).
Student 2
If lactose is absent from a cell, P1 cannot bind to LOO, so LOO is inactive. When lactose is present, it binds to P1, causing a conformational change in P1. As a result, P1 binds to LOO. When P1 is bound to LOO, LOO is active (see Figure 3).
Student 3
If lactose is absent from a cell, then galactose is also absent. If galactose is absent from a cell, then P1 cannot bind to LOO, so LOO is inactive. If lactose is present in a cell, a few lactose molecules break down into glucose and galactose without the presence of a catalyst. The galactose then binds to P1, causing a conformational change in P1. As a result, P1 binds to L OO. When P1 is bound to LOO, LOO is active (see Figure 4).
13. Based on Student 3's model, which of the following procedures will most likely inactivate an active LOO in an E. coli cell?
Your Answer is
Correct Answer is B
Explanation
Look at the second sentence of Student If galactose is absent…so LOO is inactive. So if all galactose is removed, LOO is inactive