62. Among the merodiploids of the lac operon in E. coli, which one is NOT inducible by lactose?
1. i – p o z/ i p o z
2. i – p o z/ i p- o z
3. i – p oz/ i p oz
4. i p o z/ i p o z

Introduction to Merodiploids and the Lac Operon

The lac operon in E. coli is a classic model for studying gene regulation and induction. The operon includes a set of genes involved in the metabolism of lactose, and its activity is controlled by regulatory elements. In the case of merodiploids, certain parts of the lac operon are present in duplicate — once on the chromosome and once on a plasmid. These merodiploids can help scientists determine the function of specific regulatory components.

Key Elements in the Lac Operon:

• i (lac repressor gene): Encodes the repressor protein that binds to the operator (o) and prevents transcription.

• p (promoter): Region where RNA polymerase binds to initiate transcription.

• o (operator): Binding site for the lac repressor.

• z (lacZ): Gene encoding beta-galactosidase, the enzyme that breaks down lactose.

Merodiploid Combinations and Their Inducibility:

Let’s break down each of the provided combinations:

1. i- p o z / i p o z

In this merodiploid, one copy has a null allele for the repressor gene (i-), which means that the repressor is not functional. The second copy has a functional repressor gene (i+). This results in a situation where one operon is constantly active (due to the non-functional repressor) and the other is regulated. Induction by lactose is possible because the non-functional repressor allows for expression of lacZ, but the second copy will still be regulated.

Inducibility: Inducible by lactose

2. i- p o z / i p- o z

Here, the first copy has a null repressor gene, while the second copy has a mutation in the promoter region (p-), preventing RNA polymerase from binding. As a result, the operon cannot be transcribed, even in the presence of lactose.

Inducibility: Not inducible by lactose (due to the promoter mutation in the second copy)

3. i- p o z / i p o z

This is similar to the first combination, except both operons are functional. The first copy has the non-functional repressor, while the second copy has a functional repressor. The presence of lactose will bind to the repressor, releasing it from the operator, and inducing transcription.

Inducibility: Inducible by lactose

4. i p o z / i p o z

This merodiploid has two fully functional operons. The repressor is active and can bind to the operator to prevent transcription. In the presence of lactose, the repressor is inactivated, allowing transcription to proceed. Thus, the operons are inducible by lactose.

Inducibility: Inducible by lactose

Conclusion:

The combination “i- p o z / i p- o z” is NOT inducible by lactose, because the second copy has a mutation in the promoter, preventing RNA polymerase from binding and transcribing the operon. Even in the presence of lactose, no transcription occurs due to this defect.

Final Answer:

2. i- p o z / i p- o z