BIO206H5 Lecture Notes - Lecture 17: Lac Repressor, Regulatory Sequence, Operon

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7 Dec 2016

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Review from last lecture: e. coli is a cell that demonstrates a preference for glucose compared to other sugars. The gene regulatory system is directed got make glucose the default sugar. In the absence of glucose, many di and tri and polysaccharide components can be used as a sugar source. Expression of genes: trade-off between how cells are used in glucose vs when lactose is present is absent. Genes can be negatively regulated or positively regulated. There is a protein (a repressor protein) that binds to a gene regulatory sequence near the promoter, blocks binding of rna polymerase and prevent transcription from happening. Another repressor: a protein that binds to dna but only binds to dna when it has a small molecule bound to it. (protein-small molecule complex) only in this case does it bind, blocking transcription. This is therefore relieved once the small molecule is relieved.

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In the following points, I have summarized Lac operon in Ecoli. I need to check if this is on the right track? and where does 'cleavage of disaccharides' fit into this picture?

Lactose is a disaccharide and the bacteria need specific genes to manage catabolism

1) The lacl promotor gene synthesis the repressor protein which binds to the operator gene stopping the RNA polymerase from coding the lac operon genes

2)But when lactose is present, it needs the specific genes to catabolise lactose

3) So, When allolactose (a derivative of lactose)is present, it binds to the repressor protein, the repressor releases from the operator gene

4) This allows the RNA polymerase can attach to the DNA control region

5) And drive down the operon for transcription and induction of the enzyme lacZ(beta-galactosidase) lacY and lacA which breaks lactose down into glucose and galactose

6) When the lactose is all gone, the repressor returns to the operator in a feedback loop system

When (glucose, lactose) ___________ levels inside a bacterial cell are low, cyclicAMP levels are (high, low) __________. This is because metabolizes sugars (activate, inhibit) ___________ the enzyme adenylate cyclase, and (activate, inhibit) __________ the enzyme cAMP phosphodiesterase, causing cyclic AMP levels to rise. Consequently, the CAP protein is (Activated, inhibited, unaffected) ___________ by the cyclic AMP and binds to the lac promoter region, and the CAP protein acts like a sigma factor recruiting (DNA, RNA) ___________ polymerase to the promoter. This establishes condition that may lead to the (transcription, replication) __________ of the lac operon if (allolactose, glucose, lactose) _____________ is available to bind to (lac repressor, lacZ, lacY, lacA) _________ protein keeping this protein from binding to the (lacO, lacZ. lacY, lacA) ___________ gene.

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18) In your research on gene regulation in E. coli, you produce 2 mutant strains with the same phenotype: they can grow on lactose but not on glucose. To investigate these 2 mutants further, you perform a complementation test using them and discover that the diploid combination strain can grow on glucose or lactose. From this, you can deduce that:

a.)Both mutants have a mutation in the same gene

b.)The 2 mutants have mutations in different genes
c.)Glucose and lactose metabolic pathways in E. coli utilize at least one shared gene in common
d.)At least 2 different genes are required for glucose metabolism in E. coli
e.) b) and d)

19. In your research on the lac operon, you produce a unique regulatory mutatant strain of E. Coli where the lac l gene's repressor protein binding site for the operator is normal (wild-type) but its lactose binding site no longer binds lactose. The expected phenotype of this mutant will be:

a. inability to grow on lactose only medium

b. ability to grow on lactose only medium

c. blue colony color when grown on glycerol plus x-gal medium

d. blue colony oclor when grown on lactose plus x-gal medium

e. b & d

20.

20) In your research on the lac operon, you create a diploid E.coli strain with the following genotype (+) = wild type gene, (-) = mutant gene.

I- /O+ /Z- /Y+

I+ /Oc /Z+ /Y+
Can this diploid mutant grow on medium containing lactose as the only carbon source?

a) Yes

b) No

c) Maybe, it depends on if the Oc mutation is dominant or recessive

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BIO206H5 Lecture Notes - Lecture 17: Lac Repressor, Regulatory Sequence, Operon

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