Question

Most of what we know about the lac operon of E. coli has come from genetic analysis of various mutants. See the following genotype... How much expression would you expect to see in this strain if the growth media has glucose but no lactose present?

I+P+O+Z+ CAP-

We'll use the numbers associated with transcription amounts as described in lecture and in the problem sets. H

ere is the key for these genotypes:

A “plus” superscript indicates normal functioning part of operon or repressor gene

“minus” indicates loss of function

the P region is where the RNA polymerase binds “Oc” means “operator-constitutive” mutant which means that the operator is mutated such that the repressor can’t bind “Is” means “super-repressor” which binds the operator, but doesn’t bind lactose CAPGOF indicates a gain-of-function mutation in CAP such that CAP can bind DNA even when no cAMP is present. select one a.3

b.0

c.10

d.1

Answer 1

Answer:

I am assuming that this is the mutant strain that you are asking to denote the number:

"I-P+O+Z+  CAP+"

• As indicated in the question + indicates normal function of the gene and hence promoter, operator, Z gene and CAP gene is normally functioning and there is a lack of function in the inducer gene which is denoted as I-.
• In lac operon system when no lactose is there, the inducer generates a lac repressor that binds to the operator gene and hence the transcription is stopped and when lactose is present this inducer is bound to lac repressor and operator site is free and RNA pol can bind to promoter region.
• But because in the mutant there is a non functional inducer, the lac repressor is not produced and hence operator is free and RNA pol can bind to promoter.
• But lac operon system also depends on the cAMP concentration.
• The CAP region is the gene region where the cAMP-CRP binds and allows the RNA pol to bind to the promoter region.
• This high concentration is only present when glucose level is low.
• But when glucose level is high this cAMP-CRP complex doesnt form and promoter doesnt get activated.
• Hence when there is high glucose and no lactose in the media containing this mutant, the transcription of lac operon doesnt take place.
• Hence the answer may be a.0 because the transcription is not taking place.
• This answer can be modified if you give more details on what has been taught in class regarding the number that can be assigned to each mutant.

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