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Questions regarding lac/trp operons

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Questions regarding lac/trp operons

Postby operon » Thu Mar 08, 2007 7:32 pm

Hi. I'm in a molecular biology class and I was wondering if some of you guys can verify some of my answers to two questions:

1. Wild type E. coli expresses 05 units of B-gal when grown in glucose medium and 1500 units when grown in lactose medium. How much B-gal would you expect in each of the following strains - ASSUME that RNA pol is absoultely dependant on CAP.

a) cap- i+ p+ oc z- y- / cap+ i- p+ o+ z+ y+

ANS. In presence of glucose 0.5 units of B-gal; in presence of lactose 1500 units of B-gal. Since the first strand of the merodiploid contains an inactive CAP binding site, CAP-cAMP complex won't be able to bind to the site, therefore RNA pol won't be recruited and no transcription will occur on that strand. So we are forced to look only at the second strand which is i-, but enough repressor will be made by i+ on the other strand so this will be repressed normally. We therefore get 1500 units under lactose metabolism and 0.5 units under glucose metabolism.


b) cap+ i+ p+ oc z+ y- / cap+ i- p+ o+ z+ y+

ANS. We have enough repressor in both instances but the first strand is oc, therefore the repressor tetramer can't bind to the operator region so this strand is expressed constitutively; the second strand is under negative control as normal. So we get 1500 (from strand 1) and 0.5 (from 2) = 1500.5 under glucose metabolism; and 1500+1500=3000 for lac metabolism.

QUES. Is it additive like this? Am I on the right track?

c) cap+ iS p+ o+ z+ y- / cap- i+ p+ o+ z- y+

ANS. We have production of super-repressor monomer in 1 so both operator regions are bound constitutively - so negative control doesn't work, hell we don't even get any lactose products (since it's always bound) in the presence or absence of lactose. But since we would only get B-gal from 1 anyway (2 is z-) we get 0.5 and 0.5 with both glucose and lactose metabolisms.

QUES. If strand 2 was z+ we would then get 1 for each, right? (0.5+0.5).



2. What will happen to the expression of trp when trp is removed from the following cultures? Assume attenuation accounts for 10-fold and repression accounts for 70-fold.

a) trp repressor protein mutant

ANS. With no repressor to bind to the operator, expression is only limited by attenuation. Since we are removing trp, we will be trying to produce more trp by activating the genes responsible for manufacture of trp. In this case (low trp levels - want to produce trp) attenuation will not occur (because the ribosome stalls at attenuator, antiterminator forms, transcription can proceed to trp genes) and so expression will proceed CONSTITUTIVELY until a certain point where there is enough trp to allow for a terminator to form at the attenuator (no problem inserting trp's anymore - terminator forms, RNA pol falls off and genes are not expressed). under these circumstances, repression will only account 10-fold.

QUES. He asks what happens to expression of trp under this situation, I'm telling him that repression can only be 10x and that expression would occur constitutively until attenuation can affect expression (which would be 10x).



ALRIGHT. Thanks to anyone who can reply!
operon
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Postby operon » Sat Mar 10, 2007 6:00 pm

Hi again!

Is there anyone who can help me? I have a quiz on Tuesday and I need to know if what I'm doing is indeed the correct method.
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Postby operon » Mon Mar 12, 2007 9:17 pm

Okay, forget that, consider this

I know that in the presence of glucose ALONE we still get negligible lac products (let's say 0.5 units) because there is always some allolactose present (this is the inducer, it binds to the repressor and gets off the operator, thus allowing transcription)

NOW, in an I(S) mutant, where the repressor protein is always bound to the operator, will we still see some lac products in a glucose medium?

I wouldn't think so, because the polymerase can't get past that repressor. But one of my answers says that we still get 0.5 units, what?
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