NBME 6 Block 3, Q47 - medfox87

[ArchivalUser]

The graph shows the relationship between the expression of repressor protein and the expression (transcription and translation) of a lac z gene, in wild-type Escherichia coli. Which of the following mutations in genetic elements of the lac operon could account for this relationship?

A) lac z promoter with no affinity for RNA polymerase
B) lac z protein with increased affinity for substrate
C) lac z protein with no affinity for substrate
D) Repressor with increased affinity for operator
E) Repressor with no affinity for operator

http://imageshack.us/photo/my-images/24/nbme6b3q47.jpg/

I think it's one of the no affinity ones. I'm guessing E? Can someone please explain? I've beaten lac operon to death, but for some reason it just keeps coming back.

[ArchivalUser]

This for the graph: http://img24.imageshack.us/img24/6199/nbme6b3q47.jpg

[ArchivalUser]

I agreed with E.

So lets look at the graph,

As we increase the expression of the repressor gene, the Lac z gene is still expressed at 100%.

That means the repressor is not working, lets look at the choices

A) if the promoter has no affinity for RNA pol, then transcription (expression) doesn't take place, the graph say transcription is at 100% , lets rule this out

B) lac z protein (B-galactosidase catalyzes lactose --> galactose and glucose) so increased affinity for substrate would break more lactose down into glucose, then inc glucose would inhibit cAMP, which would inhibit CAP (the enhancer), less enhancer , less transcription , lets rule this out ---

But even more this doesn't answer why the REPRESSOR isn't working

C) lac z protein with no affinity for substrate .... well, this doesn't explain why the repressor is having NO inhibitory effect on transcription (Also if lac z protein has no affinity for the substrate the minor product allolactose, is never made and the repressor is always bound ... W/E)

D) if the repressor has an increased affinity for the operator then transcription is blocked, ... the graph is showing full expression

E) if the repressor has no affinity for the operator (it wont work, it wont inhibit!!!), the lac operon will be expressed , this seems to be the best choice



I'm pretty sure that's correct .... anyone please feel free to confirm this!?

Hope that helps! best of luck with ur studying!

[ArchivalUser]

Thanks! Makes sense to me. If repressor can't bind to operator--> then there is constitutive expression of genes Z, Y, and I. Hence, Z remains flat.