1. The lac operon responds to lactose and glucose. If wild-type E. coli cells are grown in media that contains both lactose and glucose, they will consume the glucose first because: (select all that apply)
A.) | the lac repressor remains bound to lacO, preventing the lac operon from being expressed at maximal levels |
B.) | the permease (lacY) is not expressed in the presence of glucose, preventing lactose from entering the cell and being converted to allolactose, the inducer |
C.) | the CAP protein cannot bind the CAP binding site, preventing the lac operon from being expressed at maximal levels |
D.) | the lac promoter is weak, so even though lac repressor is not bound to lacO, there is far less than maximal expression of the lac operon |
2.) The small molecule that regulates CAP's ability to bind DNA: (select all that apply)
A.) | is glucose |
B.) | is an isomer of glucose |
C.) | is present at high levels in the presence of glucose |
D.) | is present at low levels in the presence of glucose |
3.) The lac operon is regulated by BOTH a repressor AND an activator because: (select all that apply)
A.) | the repressor responds to one environmental condition and the activator responds to a completely different environmental condition. |
B.) | the lac operon is an AND gate, and AND gates need to have a repressor and an activator |
C.) | the lac operon is an AND gate: it is only maximal when there is lactose AND no glucose |
D.) | operons always need one repressor and one activator or they would never be completely ON (maximal) or completely OFF (basal). |
4.) The following mutation(s) would cause the lac operon to be expressed at less than maximal levels in the presence of lactose & absence of glucose: (select all that apply)
A.) | a mutation in CAP that prevents it from binding cAMP |
B.) | a mutation in CAP that locks it in the DNA-binding conformation, independent of whether cAMP is present or not. |
C.) | a mutation in CAP that prevents it from dimerizing. |
D.) | a deletion of the adenylate cyclase gene |
Repressor is bound at lac o and prevent lac operon to express.
Cap protein is unable to bind at caps site due to lower level of cmp hence option A and C are correct
2. CAP is present in lower level binded with cAMP in higher concentration of glucose hence option C is right.
3. Operon always need a repressor or activate a protein to express operon and they are never complete not on or off hence option D is right.
4. Mutation which remain always CAP binded at cap site will result in highr expresson of operon so option B is correct.
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