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Question: problem 1c we have five lac mutants x1 x5...

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PROBLEM 1C We have five lac mutants (x1 - x5) in the bacterium E. coli about which we know that the mutation in each strain lies outside of the Z gene (the structural gene of β-galactosidase) The effects of the mutations are tested in strains that are haploid for the lac genes and in strains that are diploid for the lac region. This can be done thanks to a conjugative plasmid called F'lac, a plasmid that carries the lac portion of the E. coli genome. This plasmid encodes proteins that enable transfer of plasmid DNA to another cell. Two types of diploid strains are analyzed. In one category of diploids the lacZ gene on the F'lac plasmid is inactivated by mutation, in the other category of diploids the lacZ mutation is on the chromosome with the regulatory mutation that is being analyzed. The β−galactosidase activity in the strains was measured and is shown in Table 1B-1. IPTG is short for isopropyl-thio-β-galactoside, an inducer of β-galactosidase. For the wild type haploid strain, the β-galactosidase activity is 0.1; induced with IPTG 100

Table 1C-1 Haploid x- lacZ+ Diploid type 1 x- lacZ+ xt lacZ Diploid type 2 x- lacZ xt lacZ+ Chromosomal DNA Flac-DNA Strain Wild type IPTG IPTG 0.1 100 25 ca. 0 0.1 100 IPTG +IPTGIPTG+IPTG 100 100 100 100 100 100 0.1 0.1 0.1 100 100 100 100 25 ca. X4. 0.1 100 100 100 100 100 Remind yourself of the organization and regulation of the lac operon. lacl lacZ laclacA CAP PİO Imagine which entities can mutate. Imagine the consequence of the mutation for β-galactosidase expression Imagine the effect of adding a wild type copy of the mutated gene to the cell with the mutation. Question 1C-1. Describe the defect in each regulatory mutant

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