Fill in the blanks in this table with “yes” or “no” for each condition of lac operon regulation. The strain is wild type, with no partial diploidy. The first line is filled in for reference. Please explain your answers!
Condition |
Lactose levels high? |
lac repressor bound to operator? |
cAMP levels high? |
CAP bound to CAP binding site? |
Transcription at highest level? |
high glucose no lactose |
no |
Yes |
no |
no |
no |
no glucose high lactose |
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high glucose high lactose |
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no glucose no lactose |
Bacteria can grow on various nutrition sources. Glucose is the major source for growth.
Condition |
Lactose levels high? |
lac repressor bound to operator? |
cAMP levels high? |
CAP bound to CAP binding site? |
Transcription at highest level? |
high glucose no lactose |
no |
Yes |
no |
no |
no |
no glucose high lactose |
Yes |
no |
Yes |
no |
Yes |
high glucose high lactose |
Yes |
no |
no |
Yes |
no |
no glucose no lactose |
no |
yes |
no |
Yes |
no |
Fill in the blanks in this table with “yes” or “no” for each condition of lac...
Fill in the blanks in the following sentences by selecting from the drop-down menus. When E.coli cells are grown in the presence of high glucose and high lactose, the lac repressor is not bound to the lac operator, CAMP levels in the cell are high and CAP is mostly bound to the CAP binding site in the lac promoter. Under these conditions, transcription of the lac operon Select When E.coli cells are grown in the absence of glucose and presence...
Fill out and explain. The boxes are supposed to hae a yes or no answer. 11. Fill in the following chart for a wild type e. coli Regulated events galactosidase Cell Conditions CAMP levels CAP bound to repressor bound? transcription at highest level? high? CAP site? high glucose low lactose low glucose T high lactose high glucose high lactose low glucose low lactose
1) The lac operon is… a) Positive repressible b) Both negative inducible and positive repressible c) Positive inducible d) Negative inducible 2) An activator regulatory protein… a) Is involved in negative regulatory control b) Is a cis acting regulatory element c) Recruits RNA polymerase to the promoter d) Prevents RNA polymerase from binding the promoter 3) For the lac operon, the presence of the substrate (lactose) activates the repressor. True or False? a) true b) false 4) cAMP… a) Activates...
Please help me with this question and if you can explain why. Which of the following statements are true for bacteria grown in the absence of glucose and absence of lactose? Select all that apply. High levels of cAMP CAP bound to CAP binding site Low or no levels of allolactose O High levels of lacZYA mRNA No transcription of lacZYA Lac repressor bound to operator CAP is not bound to binding site O High levels of allolactose Lac repressor...
3. The CAP activator protein and the Lac repressor both control the Lac operon (see Figure 28-14). Fill out the table below with No expression, Low/Medium expression, or High expression to summarize when the Lac operon will be expressed in each of the three E. coli strains in the table. An example has been done for you. RNA polymerase- binding site (promoter) CAP binding site start site for RNA synthesis operator Lacz gene -80 -40 14080 nucleotide pairs Figure 08-14...
26. The lac operon in E. coli consists of genes that code for enzymes necessary for the breakdown of lactose. When lactose is absent, the operon is inactive because a repressor protein binds to a specific site in the lac operon. When lactose is present, lactose molecules bind to the repressor protein, causing the repressor protein to dissociate from the binding site. In the absence of glucose (a preferred energy source for bacteria), the protein CAP binds to a regulatory...
Imagine you are carrying out research on the lac operon. You isolate six mutations in the lac operon by measuring the amount of beta-galactosidase made in mutant cell line under three different conditions: no lactose/no glucose; lactose only; and lactose/glucose. Your results are shown in the table. Strain No Lactose, No Glucose Lactose Lactose, Glucose Wild-type None High Low Mutant 1 None None None Mutant 2 None None None Mutant 3 None Low Low Mutant 4 None Low Low Mutant...
The genotypes of the lac operon of several haploid and partial diploid E. coli strains are given. For each genotype, indicate the status of B-galactosidase synthesis as either Yes or No. Assume that no glucose is present and cAMP is rich in the growth medium. • Oc is a mutant operator that cannot be bound by the wild type repressor protein. • Is is a mutant of repressor gene and cannot bind to operator Haploid Partial Diploid IOZY IOCZY ISOZY...
A. Describe in detail the regulation of the prokaryotic lac operon for the following environment: Glucose is present, lactose present, & the repressor protein is mutated such that the allosteric site is non-functional and can no longer bind its substrate. Be sure to specify the presence or absence and location of Lac I, allolactose, cAMP, CAP, and RNA polymerase. B. Indicate whether transcription is occurring at a basal level, a high level, or not at all.
The lac operon codes for enzymes that break down lactose, but this operon is only used when needed. This includes: 1. When the repressor is bound at the operator 2.When RNA polymerase is bound to the promotor WITHOUT the CAP molecule 3.When the CAP and cAMP molecules are bound to the RNA polymerase and lactose is bound to the repressor 4.When the CAP and cAMP molecules are bound to the RNA polymerase and lactose is NOT bound to the repressor