Question

4. Monod's group figured out how the different parts of lac operon work by constructing "meroploidl E. coli cells. These have two copies of each of the lac operon genes, one in their chromosome and one on a plasmid* (actually a separate chromosome fragment introduced via conjugation), so they are "partial diploids." Many different versions of these were constructed, with different combinations of mutations in the chromosomal and/or plasmid- encoded genes. +/- indicate active/inactive gene products. O® = "constitutively active operator" (Lacl can't bind here) • Based on what you know about how the lac operon works, predict what the phenotypes of LacZ and Lacy expression in the following meroploids. The possibilities are constitutive expression ("always on”), inducible, or no expression ("always off”) Plasmid genes LacZ is.... LacY is... Chromosomal genes lacl lacO+Z+Y lacl* lacOZ Y+ lacl lacO+ZY laci lacO+ZY lacl* lacOZY lacl lacP-lacO+Z Y laci lacot lacl* lacOⓇZY lacl* lacO+Z+Y lacl lacOZY lacl lacP+lacO ZY

Answer 1

lactose is an inducer of the lac operon it inhibits the lac repressor from binding to the operator, in the absence of an inducer the lac repressor inhibits the expression of the lac operon.

lac I- lac O+ Z+ Y+/ lac I+. the operon has the mutated lac I gene but the cell has lac I+, which means the cell has the functional lac repressor, it inhibits the expression of the operon in the absence of the inducer so the expression of both Z+ and Y+ are inducible.

lac I+ lac Oc Z+Y+/lac O+ the operon has constitutive operon, the operon in the plasmid is normal but the operon is a cis-regulatory element, so operator in the plasmid cannot control the expression of both lac Z and lac Y, so the expression of the lac Z and lac Y is constitutive.

lac I+ lacO+ Z-Y+/lac I+ lac Oc Z+ Y+, the second operon has a constitutive operator and the functional genes for both beta-galactosidase and permease so the expression of both lac Z and lac Y are constitutive.

lacI+lacO+Z+Y+/lacI+lacO+Z+Y-, both operon has the normal regulatory elements the lacI+ and lacO+, so the expression of both lac operon is inducible so both lac Z and lac Y are inducible.

lacI+ lacO+ Z-Y+/lacI-lacO+ Z+Y-, both operon has the normal regulatory elements the lacI+ and lacO+, so the expression of both lac operon is inducible, the first operon has the normal Y gene and the second operon has the normal Z gene so both lac Z and lac Y are inducible.

lacI-lacP-lacO+Z+Y-/ lacI+ lac P+ lacOcZ-Y+

in the first operon, the promoter is mutated so the first operon is not transcribed and in the second operon both lac I+ and lac P+ are normal, and the operator is constitutive and this operon has no functional ac Z gene and the lac Y gene is functional so the lac Z is not expressed, and lac Y is inducible.