8. (8 pts) E. coli prefers to use of this sugar are low, adenylate cyclase produ

Question

8. (8 pts) E. coli prefers to use of this sugar are low, adenylate cyclase produces a(n) lac operon. When CAP is bound near be choose 1) as an energy source. When levels which induces change in the CAP protein, allowing it to bind near the promoter of the the promoter, and lactose is absent, the lac operon will (highly expressed, weakly expressed, completely repressed h genotype will allow the bacteria to make both lacZ and lacY in the absence of lactose? D) None of the above 10. (6 pts) A scientist targeted gene B for disruption (producing allele "b") in an agouti (AA) mouse, and produced a chimera by injecting altered germline stems cell into a black (aa), homozygous wild type B (BB) mouse. What are the possible genotypes of gametes produced by the chimera? 11. (6 pts) A plasmid vector has a gene for kanamycin resistance (Kan) and a lacZ gene, which will produce blue colonies in the presence of the chemical X-gal, if it is functional. The polylinker containing a BamHl site is in the lacZ gene. You cut your donor DNA with BamHl and mix it with BamHl digested plasmid and ligase. What will you add to the media to allow you to identify bacteria that are have received the plasmid with an insert? And how will you identify these?

Explanation / Answer

8. E coli prefers to use glucose as an energy source. When levels of this sugar are low, adenylate cyclase produces cAMP which induces a structural change in the CAP protein, allowing it to bind near the promoter of the operon. When CAP is bound near the promoter and lactose is absent,the operon will be completely repressed.(This is because, the CAP is bound because glucose levels are very low. How ever, repressor also will be bound to the operator as lactose is absent. So no transcription cannot take place, as RNA polymerase cannot bind)

9. If Oc is present, it is a constiutive operator, where transcription occurs if lactose is present or not and a repressor cannot bind to it.But in all the 3 options, beta galactosidase will be produced in the presence of lactose only,as one chromosome(in the pair) has it as non functional.gene.None of the above is correct option.

11. For screening the recombinants with BamHI, we have to add X-gal , a chromogenic substrate to the medium. If beta-galactosidase is produced (by lac Z gene)then the colonies are blue in color, as X-gal is hydrolysed by this enzyme to produce a blue pigment. This means the insert is not present in the Z gene. If the insert is right in place in lac Z gene, then beta galactosidase is not produced and colonies appear white(recombinant).We can know that these bacteria have plasmids and are recombinant.

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