GEN Exam 3-Oct 28- Regulation of Transcription in Prokaryotes (lac operon)

Allo-lactose

Allo-lactose is an effector for the Lac repressor protein that causes an allosteric shape change in the repressor. The new protein conformation can no longer perform its biological function, which is to repress transcription of the lac operon.

Select the components of the lac operon or the regulators of the lac operon that are cis-acting elements. CHOOSE ALL THAT APPLY.

CAP site operator -10 region

The highest levels of transcription of the lac operon occur when which THREE of these are present?

CAP-cAMP lactose RNA polymerase

When cAMP levels are high, glucose levels are ________and cAMP ____to CAP.

Low, Binds

Select all the statements that are true about the operator region of the lac operon

The operator is classified as a cis-acting element The operator does not change shape when a protein binds to it. The repressor protein binds to the operator when lactose is absent

An operon is

a DNA region where one promoter directs transcription of one RNA that encodes two or more completely different proteins

A mutation occurs in the lac repressor protein such that the repressor can no longer undergo an allosteric transition when allo-lactose binds. This mutation affects

ability of repressor to un-bind from the operator

Select all molecules that function as effectors

allo-lactose cyclic AMP

The product of the lacZ gene is beta-galactosidase and one of its functions is to

break a covalent bond between glucose and galactose (Another function of beta-galactosidase is to convert lactose into allo-lactose.)

cyclic AMP (cAMP)

is an effector for the Catabolite Activator Protein (CAP) that enables CAP to display a new function - binding to the CAP binding site in the lac promoter region.

Select the components of the lac operon or the regulators of the lac operon that can undergo an allosteric transition.

repressor protein CAP protein

Select the components of the lac operon or the regulators of the lac operon that are trans-acting factors. CHOOSE ALL THAT APPLY.

repressor protein CAP-cAMP

This question also asks you to compare the effects of two different mutations involving components of the lac operon. 1. One cell carries a mutation in the repressor gene (IS or super repressor) that prevents release of the repressor from the operator. 2.The second cell has a mutation in the CAP site that prevents the CAP-cAMP complex from binding. Assume that both cells are grown separately but in the same type of medium with lactose and without glucose. Which statement is true?

the amount of lac operon transcription is less in the super-repressor mutants. 1. Because there is no glucose, CAP-cAMP is bound to the CAP site but, because the repressor can't release from the operator, it doesn't matter whether lactose is present or not - the repressor simply can't undergo the required allosteric shape change that would allow it to release from the operator. In this situation, RNA polymerase can NOT transcribe the lac operon. 2. The CAP site mutation prevents CAP-cAMP from binding to boost up transcription to a very high level as would be expected in growth medium lacking glucose. The presence of lactose in the medium indicates that the repressor would not be bound to the operator and RNA polymerase would transcribe the lac operon, just not to the elevated levels that would be expected if CAP-cAMP were bound.

One cell lacks the repressor protein due to a mutation in the I gene (I-or I-minus). Another cell has a mutation in the operator sequence the prevents repressor from binding (Oc or operator constitutive). Assume that both cells are grown separately but in the same type of medium with lactose and without glucose. Which statement is true?

the amount of lac operon transcription is very similar in both mutant cells (the cell that lacks repressor protein so there is no way to prevent RNA polymerase from transcribing the lac operon. Lacks glucose=cAMP high.CAP-cAMP complex will be bound to the CAP site. CAP-cAMP increases the affinity of RNA polymerase for the lac promoter, so transcription will be high.) Second mutation reasoning (there is again no way to prevent RNA polymerase from transcribing the lac operon.Since the medium lacks glucose, cellular levels of cAMP will be high and thus the CAP-cAMP complex will be bound to the CAP site. CAP-cAMP increases the affinity of RNA polymerase for the lac promoter, so transcription will be high and should be very similar to the level of transcription in the repressor mutant.)

An E. coli cell carries an OC mutation. If both glucose and lactose are absent (perhaps the cell is using maltose as its carbon and energy source), expression of the lac operon will be

very high OC means "operator constitutive", describing a type of mutation where the operator is not recognized by the repressor protein due to sequence changes in the operator itself. As a result, the lac operon is always transcribed. It actually doesn't matter if lactose is present or not because the repressor is unable to bind to the operator under any circumstance. These facts allow you to rule out "almost non-existent" as the correct answer. Even when there is no repressor present, remember that RNA polymerase needs CAP-cAMP to be bound at the CAP site in order to really ramp up transcription of the lac operon. CAP-cAMP binding occurs when glucose is LOW, because low glucose results in plenty of cAMP, which is the effector molecular for CAP. Since the question states that glucose is low, you know that cAMP is high, that cAMP-CAP will be bound to the CAP site, and that transcription of the operon will be maximal.