Question

Arrange the following steps in lac operon induction in correct order :
A. Formation of mRNA
B. Binding of lactose to repressor
C. RNA polymerase binds to promoter
D. Transcription of structural genes
E. Removal of repressor from operator
Choose the correct answer from the options given below :

• A. \(D, E, A, B, C\)
• B. \(E, B, D, A, C\)
• C. \(C, A, E, B, D\)
• D. \(B, E, C, D, A\)

Hint:

Remember the sequence of lac operon induction: \[ \text{Inducer binds repressor} \rightarrow \text{Repressor removed} \rightarrow \text{Transcription begins} \]
• Lactose/allolactose \(\rightarrow\) Inducer
• lacZ \(\rightarrow\) \(\beta\)-galactosidase
• lacY \(\rightarrow\) Permease

Answer 1

The Correct Option is D

Concept: The lac operon in E.coli controls the metabolism of lactose. It is an inducible operon that becomes active in the presence of lactose. Important components of the lac operon include:
• Regulator gene (\(lacI\))
• Promoter
• Operator
• Structural genes (\(lacZ, lacY, lacA\)) In the absence of lactose: \[ \text{Repressor protein binds to operator} \] and blocks transcription. When lactose is present, it acts as an inducer.

Step 1: Binding of lactose to repressor. Lactose (actually allolactose) binds to the repressor protein: \[ \boxed{B} \] This changes the shape of the repressor.

Step 2: Removal of repressor from operator. After inducer binding: \[ \text{Repressor detaches from operator} \] Thus: \[ \boxed{E} \]

Step 3: RNA polymerase binds to promoter. With the operator unblocked: \[ \text{RNA polymerase can bind to promoter} \] Thus: \[ \boxed{C} \]

Step 4: Transcription of structural genes. RNA polymerase transcribes the lac structural genes: \[ lacZ,\ lacY,\ lacA \] Thus: \[ \boxed{D} \]

Step 5: Formation of mRNA. As transcription proceeds: \[ \text{mRNA is formed} \] Thus: \[ \boxed{A} \] Therefore, the correct order is: \[ \boxed{B \rightarrow E \rightarrow C \rightarrow D \rightarrow A} \] Hence, the correct answer is: \[ \boxed{(D)\ B, E, C, D, A} \]