BIOL 3305 Lecture Notes - Lecture 8: Lac Operon, Dna Binding Site, Tata Box

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6 Aug 2018

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(cid:131) can use something as a substitute instead of lactose, to make it easier to measure. (cid:120) Hard to investigate when things go up and down. Inducer binds to repressor, repressor going to be inactive, and its no longer bound to the operator. (cid:120) can now have synthesis. (cid:131) activation kicks in, high camp, no glucose. Going to produce even more lacz. (cid:120) glucose is the currency of the cell: another system that is a positive regulatory system. (cid:131) Involves camp, and involves the protein called cap, which binds camp. (cid:131) that"s going to give us positive regulation. (cid:131) camp is inversely correlated wit glucose. Rna polymerase to start transcription transcribe introns, exons, 5" utr and 3" utr), regulatory sequence, 5" and promoter proximal. 5" point of first nucleotide is +1, to the left is upstream and to the right is downstream. Some are the same, but some are different.

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1. If a bacterium with a wildtype lac operon is grown in the presence of lactose, high levels of transcription result for a while, however, eventually transcription is again down-regulated. If, on the other hand, the sugar IPTG is present (instead of lactose), transcription levels are high and remain high. What function do IPTG and lactose share? How are they differently affected by the operon product?

Lactose and IPTG are both inducers of the Lac-Operon. Lactose is a substance that the enzyme will react with (substrate), which is split or cleaved by the Beta-galactosidase. When you add lactose expressions level of lac operon will increase but as lactose is used by the beta-galactosidase expression levels of lac operon will decrease.

IPTG is a chemical cannot be split or cleaved by beta-galactosidase. The expression level of lac operon will remain the same.

Or Lactose can be easily broken down but IPTG cannot be broken down. What does expressions level mean?

2. If a lacI - bacterium is grown in the presence of the sugar X-gal, a blue dye appears; however, the concentration of X-gal decreases over time. What has happened to the X-gal?

If a lacI - bacterium is grown in the presence of the sugar X-gal, the Lacl is mutant or the gene is damaged which alters the genetic message carried by that gene. The LacI gene produces the repressor protein which switches off the operon, but since it is damaged no repressor protein is produced and the operon is switched on. The Lacl is transcribed into beta-galactosidase (lactose) The enzyme beta-galactosidase (lactose) splits or separates the X-gal which will result in decreased concentration of X-gal with time.

3. If a wildtype bacterium is grown in the presence of the sugar X-gal, transcription of the polycistronic m-RNA does not occur. What is X-gal unable to do? Without adding lactose (or genetically modifying the bacteria), how could you fix this problem

X-gal cannot form lactose, the repressor remains off and beta-galactosidase is not formed. This can be fixed by adding IPTG which can switch on the operon

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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BIOL 3305 Lecture Notes - Lecture 8: Lac Operon, Dna Binding Site, Tata Box

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