Maximal transcription of the lac operon requires

Question

Maximal transcription of the lac operon requires

BlackTom AI · Accepted Answer

CAP (catabolite activator protein) must be bound by cyclic AMP and the lac repressor must be inactivated by lactose (actually allolactose) for maximal lac operon transcription.

1) Relevant concepts
- Negative control: lac repressor binds the operator and blocks RNA polymerase; allolactose (an isomer of lactose) binds the repressor and causes it to release the operator, permitting transcription.
- Positive control (catabolite repression): high glucose lowers intracellular cAMP; when glucose is low, cAMP rises, cAMP binds CAP; the CAP–cAMP complex binds the promoter region and strongly stimulates RNA polymerase binding and transcription initiation.
Maximal expression therefore requires both repressor removal (lactose present) and CAP activation (high cAMP, i.e., low glucose).

2) Step-by-step reasoning
- Step 1: If lactose is present, some lactose is converted to allolactose, which binds the lac repressor and causes it to dissociate from the operator. This removes the block to transcription.
- Step 2: If glucose is low, intracellular cAMP concentration increases. cAMP binds CAP; the CAP–cAMP complex binds upstream of the lac promoter and recruits/stabilizes RNA polymerase.
- Step 3: With the operator free and CAP–cAMP bound, RNA polymerase efficiently initiates transcription, producing maximal levels of lac operon mRNA.

3) Intermediate “logic” steps (no numeric calculations)
- Lactose present → repressor inactive.
- Low glucose → high cAMP → CAP active.
- Both conditions true → maximal transcription.

4) Why the correct answer is correct
“The presence of cyclic AMP and lactose” provides both positive activation (CAP–cAMP) and removal of negative regulation (allolactose-mediated repressor release), which together produce maximal transcription.

5) Why other options are incorrect
- High levels of glucose: suppresses cAMP, reducing CAP activation, so transcription is low (catabolite repression).
- Removal of the operator region: would prevent repressor binding but does not supply CAP activation; also not the physiological regulatory mechanism and could lead to unregulated expression but not necessarily the maximal regulated response.
- Activation of the repressor protein: would increase repression and decrease transcription.
- Presence of cyclin and CDK: eukaryotic cell-cycle regulators, irrelevant to bacterial lac operon regulation.

Conclusion: Maximal transcription requires cyclic AMP (for CAP activation) and lactose (to inactivate the repressor). This matches the provided answer: the presence of cyclic AMP and lactose.

Maximal transcription of the lac operon requires单项选择题

类似问题

How does the catabolite repression mechanism, involving the cAMP-CAP complex, modulate the expression of the lac operon when both glucose and lactose are present?

In the presence of lactose but absence of glucose, what happens to cAMP levels and the lac operon?

A researcher grows E. coli in minimal medium containing both glucose and lactose and observes that the lac operon is not highly expressed. Which explanation best accounts for this observation?

A mutation in the lacZ gene alters β-galactosidase so it makes more allolactose compared with a normal β-galactosidase enzyme. How would this mutation affect the expression of the lac operon?

The binding of ___________ to the lac repressor disables the lac repressor to bind to the site, thereby transcription.

On the chromosome, CAP is mutated such that it no longer produces a functional CAP protein, and the operator is mutated such that it no longer binds to the repressor. On the plasmid, lacI is mutated so that it no longer produces a functional repressor protein. Lactose and glucose are both absent.

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